^
THE
AMERICAN HOUSE CARPENTER.
A TREATISE
ON THE
ART OF BUILDING.
COMPRISING
STYLES OF ARCHITECTURE, STRENGTH OF MATERIALS,
B D
THE THEORY AND PRACTICE OF THE CONSTRUCTION OF FLOORS, FRAMED
GIRDERS, ROOF TRUSSES, ROLLED-IRON BEAMS, TUBULAR-IRON
GIRDKRS, CAST-IRON GIRDERS, STAIRS, DOORS,
WINDOWS, MOULDINGS, AND CORNICES;
A COMPEND OF MATHEMATICS.
A MANUAL FOR THE PRACTICAL USE OF
ARCHITECTS, CARPENTERS, STAIR-BUILDERS,
AND OTHERS.
EIGHTH EDITION,
REWRITTEN AND ENLARGED.
BY
R. G. HATFIELD, ARCHITECT,
l\
LATE FELLOW OF THE AMERICAN INSTITUTE OF ARCHITECTS, MEMBER OF THE AMERICAN
SOCIETY OF CIVIL ENGINEERS, ETC.
AUTHOR OF "TRANSVERSE STRAINS."
EDITED BY O. P. HATFIELD, F.A.I.A., ARCHITECT.
NINTH EDITION.
NEW YORK:
JOHN WILEY & SONS, 15 ASTOR PLACE
1883.
^/.o
\
COPYRIGHT 1880,
BY THE ESTATE OF R. G. HAITI ELD.
PREFACE.
SINCE the publication of the first edition of this work, six subsequent
editions have been issued ; but, although from time to time many additions
to its pages and revisions of its subject-matter have been made, still its sev-
eral issues have always been printed substantially from the original stereotype
plates. In this edition, however, the book has been extensively remodelled
and expanded, the greater portion of it rewritten, and the whole put in a new
dress by being newly set up in type uniform in style with that of the late
author's recent work, Transverse Strains. To this revision — a labor of love to
him — he devoted all the time he could spare from his other pressing engage-
ments for a year or more, and by close and arduous application brought the
book to a successful termination, notwithstanding the engrossing nature of
his customary business avocations. Although essentially an elementary work,
and intended originally for a class of minds not generally favored with oppor-
tunities for securing a very extended form of education, either in the store of
* <r * N x* N:
information acquired or in the discipline of mind which culture confers, still
it has been his aim to embody in its pages so complete and exhaustive a treat-
ment of the various subjects discussed, and so practical and useful a collection
of data and the rules governing their application, as to make it also not un-
worthy the attention of those who have been more highly favored in that
respect.
In all the various trades connected with building it is the intelligent
workman that commands the greatest respect, and who receives in all cases
the highest remuneration. As apprentice, journeyman, and master-builder,
his course is upward and onward, and success crowns his efforts in all that he
undertakes. There is a kind of freemasonry in the very air that surrounds
the skilful, intelligent man, that gives him a pass at once into the appreciation
and recognition of all those whose regard is valuable. We admire and respect
the plodding toil of the honest, patient laborer, whose humble task may tax
his muscles though not his mind, but we yield a deeper homage to the skilful
hand and tutored eye that accomplish wonders in art and science through per-
severance in aspiring studies. It was to excite in the minds of workmen like
these an ambition to excel in their calling, and to point out to them the surest
path to -that consummation, that the preparation of this volume was under-
taken ; that all its tendencies are in that direction, and that it cannot well fail
11 PREFACE.
of its purpose when judiciously used, must be the conviction of all who will
take the trouble to examine its pages.
In the first part of the book matters more particularly relating to building
are treated of. The first section is in the nature of an introduction, serving
by its historical references to excite an interest in the general subject, while
in the second are presented the methods of erecting edifices in accordance
with the acknowledged principles of sound construction. In the remaining
sections of Part I. the several well-defined branches of house-building, as
stairs, doors and windows, etc., are illustrated and explained. In the second
part the more useful rules and simple problems of mathematics are reduced
to an easily acquired form, and adapted to the necessities of the ordinary
workman. By studying the latter, the young mechanic may not only improve
and strengthen his mind, but grow more self-reliant daily, demonstrating in
his own experience that scientific knowledge gives power. By carefully
studying this part of the book he will see how easy it is to acquire the knowl-
edge of solving problems by signs and symbols, commonly called Algebra
(although looked upon by the uninitiated as almost incomprehensible), and
thus find it easy to understand all the illustrations of the various subjects
wherein those condensed forms of expression are used. Useful problems in
geometry, described in simple^ language, and hints upon the subject of draw-
ing and shading, are also to be found in Part II. A glossary of architectural
terms and many useful tables are provided in the Appendix, and finally, an
Index is added to aid in referring to special subjects. The full-plate illustra-
tions are inserted to make it attractive to the general reader, and at the same
time to serve as explanatory of the historical portion of the volume.
It will not be denied that the class of information herein furnished is one
of the most instructive and useful that can.be presented to the practical mind
of a workingman, or to any mind engaged in mechanical pursuits. The im-
press stamped upon it by the author's peculiar line of study is not to be
effaced, but this has given it characteristics of originality and strength not
to be found in a mere compilation.
THE EDITOR.
NEW YORK, 31 Pine Street,
January 6, 1880.
CONTENTS.
For a Table of Contents more in detail, see p. 613, and for Index, see p. 657.
PART I.
PAGE.
SECTION I. — Architecture 5
II. — Construction 57
III. — Stairs 240
" IV. — Doors and Windows 315
" V. — Mouldings and Cornices 323
PART II.
SECTION VI.— Geometry 347
VII. — Ratio or Proportion 366
" VIII. — Fractions 378
IX. — Algebra 392
X.— Polygons 439
XL— The Circle 468
'• XII.— The Ellipse 481
" XIII.— The Parabola 492
" XIV.— Trigonometry 5IQ
XV.— Drawing 53^
" XVI. — Practical Geometry 544
" XVII.— Shadows..... S9&
Table of Contents in detail . . 6l3
iv .CONTENTS.
APPENDIX.
Glossary 627
Table of Squares, Cubes and Roots 638
Rules for the Reduction of Decimals 647
Table of Circles 649
Table showing the Capacity'of Wells, Cisterns, etc 653
Table of the Weights of Materials 654
Index 657
((UNIVERSITY
UNIVERSITY
PART I.
SECTION L— ARCHITECTURE.
ART. I. — Building Defined. — Building" and Architecture
are technical terms by some thought to be synonymous ;
but there is a distinction. Architecture has been defined to
be— "the art of building;" but more correctly it is — "the
art of designing and constructing buildings, in accordance
with such principles as constitute stability, utility, and
beauty." The literal signification of the Greek word arclii-
tecton, from which the word architect is derived, is chief-
carpenter ; and the architect who designs and builds well
may truly be considered the chief builder. Of the three
classes into which architecture has £>een divided — viz., Civil,
Military, and Naval — the first is that which refers to the
construction of edifices known as dwellings, churches, and
other public buildings, bridges, etc., for the accommodation
of civilized man— and is the subject of the remarks which
follow. .
2.— Antique Buildings; Tower of Babel.— Building is
one of the most ancient of the arts : the Scriptures inform
us of its existence at a very early period. Cain, the son of
Adam, " builded a city, and called the name of the city
after the name of his son, Enoch ;" but of the peculiar style
or manner of building we are not informed. It is presumed
that it was not remarkable for beauty, but that utility and
perhaps stability were its characteristics. Soon after the
deluge — that memorable event, which removed from ex-
istence all traces of the works of man— the Tower of Babel
6 ARCHITECTURE.
was commenced. This was a work of such magnitude that
the gathering of the materials, according to some writers,
occupied three years; the period from its commencement
until the work was abandoned was twenty-two years ; and
the bricks were like blocks of stone, being twenty feet long,
fifteen broad, and seven thick. Learned men have given it
as their opinion that the tower in the temple of Belus at
Babylon was the same as that which in the Scriptures is
called the Tower of Babel. The tower of the temple of
Belus was square at its base, each side measuring one
furlong, and consequently half a mile in circumference. Its
form was that of a pyramid, and its height was 660 feet. It
had a winding passage on the outside from the base to the
summit, which was wide enough for two carriages.
3. — Ancient Cities and Monuments. — Historical accounts
of ancient cities, such as Babylon, Palmyra, and Nineveh of
the Assyrians ; Sidon, Tyre, Aradus, and Serepta of the
Phoenicians; and Jerusalem, with its splendid temple, of
the Israelites — show that architecture among them had
made great advances. Ancient monuments of the art are
found also among other nations ; the subterraneous temples
of the Hindoos upon the islands Elephanta and Salsetta ;
the ruins of Persepolis in Persia ; pyramids, obelisks, tem-
ples, palaces, and sepulchres in Egypt — all prove that the
architects of those early times were possessed of skill and
judgment highly cultivated. The principal characteristics
of their works are gigantic dimensions, immovable solidity,
and, in some instances, harmonious splendor. The extra-
ordinary size of some is illustrated in the pyramids of Egypt.
The largest of these stands not far from the city of Cairo :
its base, which is square, covers about ir£ acres, and its
height is nearly 500 feet. The stones of which it is built
are immense — the smallest being full thirty feet long.
4. — Architecture in Greece. — Among the Greeks, archi-
tecture was cultivated as a fine art. Dignity and grace
were added to stability and magnificence. In the Doric
order, their first style of building, this is fully exemplified.
Phidias, Ictinus, and Calicrates are spoken of as masters in
GRECIAN AND ROMAN BUILDINGS. 7
the art at this period : the encouragement and support of
Pericles stimulated them to a noble emulation. The beauti-
ful temple of Minerva, called the Parthenon, erected upon
the acropolis of Athens, the Propyleum, the Odeum, and
others, were lasting monuments of their success. The Ionic
and Corinthian orders were added to the Doric, and many
magnificent edifices arose. These exemplified, in their
chaste proportions, the elegant refinement of Grecian taste.
Improvement in Grecian architecture continued to advance
until perfection seems to have been attained. The speci-
mens which have been partially preserved exhibit a com-
bination of elegant proportion, dignified simplicity, and
majestic grandeur. Architecture among the Greeks was at
the height of its glory at the period immediately preceding
the Peloponnesian war ; after which the art declined. An
excess of enrichment succeeded its former simple grandeur ;
yet a strict regularity was maintained amid the profusion of
ornament. After the death of Alexander, 323 B.C., a love
of gaudy splendor increased : the consequent decline of the
art was visible, and the Greeks afterwards paid but little
attention to the science.
5. — Architecture in Rome. — While the Greeks illustrated
their knowledge of architecture in the erection of their
temples and other public buildings, the Romans gave their
attention to the science in the construction of the many
aqueducts and sewers with which Rome abounded ; build-
ing no such splendid edifices as adorned Athens, Corinth,
and Ephesus, until about 200 years B.C., when their inter-
course with the Greeks became more extended. Grecian
architecture was introduced into Rome by Sylla ; by whom,
as also by Marius and Caesar, many large edifices were
erected in various cities of Italy. But under Csesar Augus-
tus, at about the beginning of the Christian era, the art
arose to the greatest perfection it ever attained in Italy.
Under his patronage Grecian artists were encouraged, and
many emigrated to Rome. It was at about this time that
Solomon's temple at Jerusalem was rebuilt by Herod — a
Roman. This was 46 years in the erection, and was most
probably of ' the Grecian style of building— perhaps of the
CATHEDRAL OF NOTRE
THE GOTHS AND VANDALS. 9
provement ; but very soon after his reign the art began
rapidly to decline, as particularly evidenced in the mean
and trifling character of the ornaments.
7. — Architecture Debased. — The Goths and Vandals
overran Italy, Greece, Asia, and Africa, destroying most
of their works of ancient architecture. Cultivating no art
but that of war, these savage hordes could not be expected
to take any interest in the beautiful forms and proportions
of their habitations. From this time architecture assumed
an entirely different aspect. The celebrated styles of Greece
were unappreciated and forgotten ; and modern architec-
ture made its first appearance on the stage of existence.
The Goths, in their conquering invasions, gradually ex-
tended it over Italy, France, Spain, Portugal, and Ger-
many, into England. From the reign of Galienus may be
reckoned the total extinction of the arts among the Romans.
From this time until the sixth or seventh century, architec-
ture was almost entirely neglected. The buildings which
were erected during this suspension of the arts were very
rude. Being constructed of the fragments of the edifices
which had been demolished by the Visigoths in their unre-
strained fury, and the builders being destitute of a proper
knowledge of architecture, many sad blunders and exten-
sive patch-work might have been seen in their construction
— entablatures inverted, columns standing on their wrong
ends, and other ridiculous arrangements characterized their
clumsy work. The vast number of columns which the ruins
around them afforded they used as piers in the construction
of arcades — which by some is thought, after having passed
through various changes, to have been the origin of the
plan of the Gothic cathedral. Buildings generally, which
are not of the classical styles, and which were erected after
the fall of the Roman empire, have by some been indiscrim-
inately included under the term Gothic. But the changes
which architecture underwent during the Mediaeval age
show that there were then several distinct modes of building.
8. — The O§trogoths. — Theodoric, a friend of the arts,
who reigned in Italy from A.D. 493 to 525, endeavored to
10 ARCHITECTURE.
restore and preserve some of the ancient buildings; and
erected others, the ruins of which are still seen at Verona
and Ravenna. Simplicity and strength are the character-
istics of the structures erected by him ; they are, however,
devoid of grandeur and elegance, or fine proportions.
These are properly of the GOTHIC style ; by some called
the old Gothic, to distinguish it from the pointed Gothic.
9. — The Lombard*, who ruled in Italy from A.D. 568,
had no taste for architecture nor respect for antiquities.
Accordingly, they pulled down the splendid monuments of
classic architecture which they found standing, and erected
in their stead huge buildings of stone which were greatly
destitute of proportion, elegance, or utility — their charac-
teristics being scarcely anything more than stability and
immensity combined with ornaments of a puerile character.
Their churches were decorated with rows of small columns
along the cornice of the pediment, small doors and win-
dows with circular heads, roofs supported by arches having
arched buttresses to resist their thrust, and a lavish display
of incongruous ornaments. This kind of architecture is
called the LOMBARD style, and was employed in the seventh
century in Pavia, the chief city of the Lombards ; at which
city, as also at many other places, a great many edifices
were erected in accordance with its peculiar forms.
10. — Tlic Byzantine Architects, of Byzantium, Constan-
tinople, erected many spacious edifices; among which are
included the cathedrals of Bamberg, Worms, and Mentz,
and the most ancient part of the minster at Strassburg ; in
all of these they combined the classic styles with the crude
Lombardian. This style is called the LOMBARD-BYZANTINE.
To the last style there were afterwards added cupolas sim-
ilar to those used in the East, together with numerous slen-
der pillars with elaborate capitals, and the many minarets
which are the characteristics of the proper Byzantine, or
Oriental style.
H. — The Moor*. — When the Arabs and Moors destroyed
the kingdom of the Goths, the arts and sciences were mostly
E LI
DIVERSITY)
>, ^ K /
MOSQUE AT CAIRO.
THE MEDLEVAL STYLES. II
in possession of the Musselmen-conquerors ; at which time
there were three kinds of architecture practised ; viz. : the
Arabian, the Moorish, and the Lombardian. The ARABIAN
style was formed from Greek models, having circular arches
added, and towers which terminated with globes and mina-
rets. The MOORISH is very similar to the Arabian, being
distinguished from it by arches in the form of a horseshoe.
It originated in Spain in the erection of buildings with the
ruins of Roman architecture, and is seen in all its splendor
in the ancient palace of the Mohammedan monarchs at
Grenada, called the Alhambra, or red-house. The style which
was originated by the Visigoths in Spain by a combination
of the Arabian and Moorish styles, was introduced by Charle-
magne into Germany. On account of the changes and im-
provements it there underwent, it Was, at about the I3th or
I4th century, termed the German or romantic style. It is ex-
hibited in great perfection in the towers of the minster of
Strassburg, the cathedral of Cologne and other edifices.
The most remarkable features of this lofty and aspiring style
are the lancet or pointed arch, clustered pillars, lofty towers,
and flying buttresses. It was principally employed in eccle-
siastical architecture, and in this capacity introduced into
France, Italy, Spain, and England.
12. — Ttie Architecture of England: is divided into the
Norman, the Early-English, the Decorated, and the Perpendic-
ular styles. The Norman is principally distinguished by
the character of its ornaments — the chevron, or zigzag, being
the most common. Buildings in this style were erected in
the 1 2th century. The Early-English is celebrated for the
beauty of its edifices, the chaste simplicity and purity of
design which they display, and the peculiarly graceful char-
acter of its foliage. This style is of the isth century. The
Decorated style, as its name implies, is characterized by a
great profusion of enrichment, which consists principally of
the crocket, or feathered-ornament, and ball-flower. It was
mostly in use in the Hth century. The Perpendicular style,
which dates from the I5th century, is distinguished by its
high towers, and parapets surmounted with spires similar in
number and grouping to oriental minarets.
12 ARCHITECTURE.
13.— Architecture Progresiiive.— The styles erroneously
termed Gothic were distinguished by peculiar characteris-
tics as well as by different names. The first symptoms of a
desire to return to a pure style in architecture, after the
ruin caused by the Goths, was manifested in the character
of the art as displayed in the church of St. Sophia at Con-
stantinople, which was erected by Justinian in the 6th
century. The church of St. Mark at Venice, which arose
in the loth or nth century, is a most remarkable building;
a compound of many of the forms of ancient architecture.
The cathedral at Pisa, a wonderful structure for the age,
was erected by a Grecian architect in 1016. The marble
with which the walls of this building were faced, and of
which the four rows of columns that support the roof are
composed, is said to be of an excellent character. The
Campanile, or leaning-tower as it is usually called, was
erected near the cathedral in the I2th century. Its inclina-
tion is generally supposed to have arisen from a poor foun-
dation ; although by some it is said to have been thus con-
structed originally, in order to inspire in the minds of the
beholder sensations of sublimity and awe. In the I3th cen-
tury, the science in Italy was slowly progressing ; many fine
churches were erected, the style of which displayed a de-
cided advance in the progress towards pure classical archi-
tecture. In other parts of Europe, the Gothic, or pointed
style was prevalent. The cathedral at Strassburg, designed
by Irwin Steinbeck, was erected in the I3th and I4th cen-
turies. In France and England during the I4th century,
many very superior edifices were erected in this style.
14-. — Architecture in Italy. — In the I4th and 1 5th cen-
turies, architecture in Italy was greatly revived. The mas-
ters began to study the remains of ancient Roman edifices ;
and many splendid buildings were erected, which displayed
a purer taste in the science. Among others, St. Peter's of
Rome, which was built about this time, is a lasting monu-
ment of the architectural skill of the age. Giocondo, Mi-
chael Angelo, Palladio, Vignola, and other celebrated archi-
tects, each in their turn, did much to restore the art to its
INTERIOR OF ST
SOPHIA, CONSTANTINOPLE.
-gjpl Lift
Y^
UNIVERSITY]
ORIGIN OF STYLES. 13
former excellence. In the edifices which were erected under
their direction, however, it is plainly to be seen that they
studied not from the pure models of Greece, but from the
remains of the deteriorated architecture of Rome. The
high pedestal, the coupled columns, the rounded pediment,
the many curved-and-twisted enrichments, and the convex
frieze, were unknown to pure Grecian architecture. Yet
their efforts were serviceable in correcting, to a good de-
gree, the very impure taste that had prevailed since the over-
throw of the Roman empire.
15. — The Renaissance. — The Italian masters and numer-
ous artists who had visited Italy for the purpose, spread the
Roman style over various countries of Europe; which was
gradually received into favor in place of the pointed Gothic.
This fell into disuse ; although it has of late years been
again cultivated. It requires a building of great magnitude
and complexity for a perfect display of its beauties. In
America, the pure Grecian style was at first more or less
studied ; and perhaps the simplicity of its principles would
be better adapted to a republican country than the more
intricate mediaeval styles ; yet these, during the last quarter
of a century, have been extensively studied, and now wholly
supersede the Grecian styles.
16. — Style§ of Arehiteeture. — It is generally acknowl-
edged that the various styles in architecture were the results
of necessity, and originated in accordance with the different
pursuits of the early inhabitants of the earth ; and were
brought by their descendants to their present state of per-
fection, through the propensity for imitation and desire of
emulation which are found more or less among- all nations.
Those that followed agricultural pursuits, from being em-
ployed constantly upon the same piece of land, needed a
permanent residence, and the wooden hut was the offspring
of their wants ; while the shepherd, who followed his flocks
and was compelled to traverse large tracts of country for
pasture, found the tent to be the most portable habitation ;
again, the man devoted to hunting and fishing — an idle and
vagabond way of living — is naturally supposed to have been
14 ARCHITECTURE.
content with the cavern as a place of shelter. .The latter is
said to have been the origin of'the Egyptian style; while
the curved roof of Chinese structures gives a strong indica-
tion of their having had the tent for their model ; and the
simplicity of the original style of the Greeks (the Doric)
shows quite conclusively, as is generally conceded, that its
original was of wood. The pointed, or ecclesiastical style,
is said to have originated in an attempt to imitate the bower,
or grove of trees, in which the ancients performed their idol-
worship. But it is more probably the result of repeated
scientific attempts to secure real strength with apparent
lightness ; thus giving a graceful, aspiring effect.
17.— Order§: or styles, in architecture are numerous;
and a knowledge of the peculiarities of each is important to
the student in the art. An ORDER, in architecture, is com-
posed of three principal parts, viz. : the Stylobate, the Col-
umn, and the Entablature. This appertains chiefly to the
classic styles.
18. — The Stylobate: is the substructure, or basement,
upon which the columns of an order are arranged. In
Roman architecture — especially in the interior of an edi-
fice— it frequently occurs that each column has a separate
substructure ; this is called a pedestal. If possible, the ped-
estal should be avoided in all cases ; because it gives to the
column the appearance of having been originally designed
for a small building, and afterwards pieced out to make it
long enough for a larger one.
19. — The Column : is composed of the base, shaft, and
capital.
20. — Tlie Entablature: above and supported by the
columns, is horizontal ; and is composed of the architrave,
frieze, and cornice. These principal parts are again divided
into various members and mouldings.
21. — The Base: of a column is so called from basis, a
foundation or footing.
^t° ^"4j>2\
7A >
UNIVERSITY)
INTERIOR OF ST. STEPHENS, PARIS.
PARTS OF AN ORDER. 15
22.— The Shaft: the upright part of a column standing
upon the base and crowned with the capital, is from shafto,
to dig — in the manner of a well, whose inside is not unlike
the form of a column.
23. — The Capital : from kephale or caput, the head, is the
uppermost and crowning part of the column.
24. — The Architrave : from archi, chief or principal,
and trabs, a beam, is that part of the entablature which lies
in immediate connection with the column.
25. — The Frieze: from fibron, a fringe or border, is that
part of the entablature which is immediately above the
architrave and beneath the cornice. It was called by some
of the ancients zophorus, because it was usually enriched
with sculptured animals.
26. — The Corniee: from corona, a crown, is the upper
and projecting part of the entablature — being also the upper-
most and crowning part of the whole order.
27. — The Pediment: above the entablature, is the tri-
angular portion which is formed by the inclined edges of
the roof at the end of the building. In Gothic architecture,
the pediment is called a gable.
28. — The Tympanum: is the perpendicular triangular
surface which is enclosed by the cornice of the pediment.
29. — The Attic : is a small order, consisting of pilasters
and entablature, raised above a larger order, instead of a
pediment. An attic story is the upper story, its windows
being usually square.
30. — Proportions in an Order. — An order has its several
members proportioned to one another by a scale of 60 equal
parts, which are called minutes. If the height of buildings
were always the same, the scale of equal parts would be a
fixed quantity— an exact number of feet and inches. But as
buildings are erected of different heights, the column and
1 6 ARCHITECTURE.
its accompaniments are required to be of different dimen-
sions. To ascertain the scale of equal parts, it is necessary
to know the height to which the whole order is to be
erected. This must be divided by the number of diameters
which is directed for the order under consideration. Then
the quotient obtained by such division is the length of the
scale of equal parts — and is, also, the diameter of the column
next above the base. For instance, in the Grecian Doric
order the whole height, including column and entablature,
is 8 diameters. Suppose now it were desirable to construct
an example of this order, forty feet high. Then 40 feet
divided by 8 gives 5 feet for the length of the scale ; and
this being divided by 60, the scale is completed. The up-
right columns of figures, marked H and P, by the side of
the drawings illustrating the orders, designate the height
and the projection of the members. The projection of each
member is reckoned from a line passing through the axis of
the column, and extending above it to the top of the entab-
lature. The figures represent minutes, or 6oths, of the
major diameter of the shaft of the column.
31. — Grecian Styles.— The original method of building
among the Greeks was in what is called the Doric order :
to this were afterwards added the Ionic and the Corinthian.
These three were the only styles known among them. Each
is distinguished from the other two by not only a peculiar-
ity of some one or more of its principal parts, but also by a
particular destination. The character of the Doric is robust,
manly, and Herculean-like ; that of the Ionic is more deli-
cate, feminine, matronly; while that of the Corinthian is
extremely delicate, youthful, and virgin-like. However
they may differ in their general character, they are alike
famous for grace and dignity, elegance and grandeur, to a
high degree of perfection.
32 — The Doric Order: (Fig. 2,) is so ancient that its
origin is unknown— although some have pretended to have
discovered it. But the most general opinion is, that it is
an improvement upon the original wooden buildings of the
FANCIFUL ORIGIN OF THE DORIC. I/
Grecians. These no doubt were very rude, and perhaps
not unlike the following figure.
FIG i. — SUPPOSED ORIGIN OF DORIC TEMPLE.
The trunks of trees, set perpendicularly to support the
roof, may be taken for columns ; the tree laid upon the
tops of the perpendicular ones, the architrave ; the ends
of the cross-beams which rest upon the architrave, the
triglyphs ; the tree laid on the cross-beams as a support for
the ends of the rafters, the bed-moulding of the cornice ; the
ends of the rafters which project beyond the bed-moulding,
the mutules ; and perhaps the projection of the roof in
front, to screen the entrance from the weather, gave origin
to the portico.
The peculiarities of the Doric order are the triglyphs —
those parts of the frieze which have perpendicular channels
cut in their surface ; the absence of a base to the column —
as also of fillets between the flutings of the column ; and the
plainness of the capital. The triglyphs should be so dis-
posed that the width of the metopes — the space between
the triglyphs — shall be equal to their height.
33. — The Intercolumniation : or space between the col-
umns, is regulated by placing the centres of the columns
under the centres of the triglyphs — except at the angle of
the building ; where, as may be seen in Fig. 2, one edge of
18
ARCHITECTURE.
FIG. 2. — GRECIAN DORIC.
PECULIARITIES OF THE DORIC. 19
the triglyph must be over the centre of the column.*
Where the columns are so disposed that one of them stands
beneath every other triglyph, the arrangement is called
mono-triglyph and is most common. When a column is
placed beneath every third triglyph, the arrangement is
called diastyle ; and when beneath every fourth, arceostyle.
This last style is the worst, and is seldom adopted.
34-.— The Doric Order: is suitable for buildings that
are destined for national purposes, for banking-houses, etc.
Its appearance, though massive and grand, is nevertheless
rich and graceful. The Patent Office at Washington, and
the Treasury at New York, are good specimens of this
order.
35. — The Ionic Order. (Fig. 3.) — The Doric was for
some time the only order in use among the Greeks. They
gave their attention to the cultivation of it, until perfection
seems to have been attained. Their temples were the prin-
* GRECIAN DORIC ORDER. When the width to be occupied by the whole front
is limited, to determine the diameter of the column.
The relation between the parts may be expressed thus :
_ 60 a
~~ ~d(b '+ c) + (60 — c)
Where a equals the width in feet occupied by the columns, and their inter-
columniations taken collectively, measured at the base ; b equals the width
of the metope, in minutes ; c equals the width of the triglyphs in minutes ; d
equals the number of metopes, and x equals the diameter in feet.
Example. — A front of six columns — hexastyle — 61 feet wide ; the frieze
having one triglyph over each intercolumniation, or mono-triglyph. In this
case, there being five intercolumniations and two metopes over each, therefore
there are 5 x 2 = 10 metopes. Let the metope equal 42 minutes and the
triglyph equal 28. Then a = 61 ; b = 42 ; c = 28 ; and d = 10 ; and the formula
above becomes
60 x 61 60 x 61 3660
x — —. --- -— - -- - - ST = - = - — = 5 feet = the d lameter
10(42 + 28) + (60 — 28) 10x70 + 32 732
required.
Example. — An octastyle front, 8 columns, 184 feet wide, three metopes
over each, intercolumniation, 21 in all, and the metope and triglyph 42 and
28, as before. Then
l84 - = H212 = 7.35-rigir feet = the diameter required.
21 (42 + 28) + (60 - 28) 1502
20 ARCHITECTURE.
cipal objects upon which their skill in the art was displayed ;
and as the Doric order seems to have been well fitted, by its
massive proportions, to represent the character of their
male deities rather than the female, there seems to have
been a necessity for another style which should be emble-
matical of feminine graces, and with which they might
decorate such temples as were dedicated to the goddesses.
Hence the origin of the Ionic order. This was invented,
according to historians, by Hermogenes of Alabanda ; and
he being a native of Caria, then in the possession of the
lonians, the order was called the Ionic.
The distinguishing features of this order are* the volutes
or spirals of the capital ; and the dentils among the bed-
mouldings of the cornice: although in some instances
dentils are wanting. The volutes are said to have been
designed as a representation of curls of hair on the head of
a matron, of whom the whole column is taken as a sem-
blance.
The Ionic order is appropriate for churches, colleges,
seminaries, libraries, all edifices dedicated to literature and
the arts, and all places of peace and tranquillity. The front
of the Custom-House, New York City, is a good specimen
of this order.
36. — The Intercolumniation : of this and the other
orders — both Roman and Grecian, with the exception of
the Doric — are distinguished as follows. When the interval
is one and a half diameters, it is called pycnostyle, or columns
thick-set; when two diameters, systyle ; when two and a
quarter diameters, eustyle ; when three diameters, diastyle ;
and when more than three diameters, arceostyle, or columns
thin-set. In all the orders, when there are four columns in
one row, the arrangement is called tetrastyle ; when there
are six in a row, hcxastyle ; and when eight, octastyle.
37.— To Describe the Ionic Volute.— Draw a perpen-
dicular from a to s (Fig. 4), and make a s equal to 20 min.
or to $ of the whole height, a c ; draw s o at right angles to
s a, and equal to I £ min. ; upon o, with 2^ min. for radius,
PROPORTIONS OF GRECIAN IONIC.
v
FIG. 3.— GRECIAN IONIC.
22
ARCHITECTURE.
describe the eye of the volute ; about o, the centre of the
eye, draw the square, r t i 2, with sides equal to half the
diameter of the eye, viz. 2j min., and divide it into 144 equal
parts, as shown at Fig. 5. The several centres in rotation are
at the angles formed by the heavy lines, as figured, i, 2, 3,
4, 5, 6, etc. The position of these angles is determined by
commencing at the point, i, and making each heavy line one
part less in length than the preceding one. No. i is the
FIG. 4. — IONIC VOLUTE.
THE IONIC VOLUTE. 23
centre for the arc a b (Fig. 4 ;) 2 is the centre for the arc
be; and so on to the last. The inside spiral line is to be
described from the centres, x, x, x, etc. (Fig. 5), being the
centre of the first small square towards the middle of the
eye from the centre for the outside arc. The breadth of the
fillet at aj is to be made equal to 2T3¥ min. This is for a spiral
of three revolutions ; but one of any number of revolutions,
as 4 or 6, may be drawn, by dividing of (Fig. 5) into a cor-
responding number of equal parts. Then divide the part
nearest the centre, o, into two parts, as at h ; join o and i,
also o and 2 ; draw h 3 parallel to o i, and h 4 parallel to o
FIG. 5. — EYE OF VOLUTE.
2 ; then the lines o i, o 2, // 3, h 4 will determine the length
of the heavy lines, and the place of the centres. (See Art.
288.)
38.— The Corinthian Order : (Fig. /,) is in general like
the Ionic, though the proportions are lighter. The Corin-
thian displays a more airy elegance, a richer appearance ;
but its distinguishing feature is its beautiful capital,
is generally supposed to have had its origin in the capitals
24 ARCHITECTURE.
of the columns of Egyptian temples, which, though not ap-
proaching it in elegance, have yet a similarity of form with
the Corinthian. The oft-repeated story of its origin which
is told by Vitruvius — an architect
who flourished in Rome in the days
of Augustus Caesar — though pretty
generally considered to be fabu-
lous, is nevertheless worthy of be-
ing again recited. It is this : A
young lady of Corinth was sick, and
finally died. Her nurse gathered
, into a deep basket such trinkets and
keepsakes as the lady had been
fond of when alive, and placed them upon her grave, cover-
ing the basket with a flat stone or tile, that its contents
might not be disturbed. The basket was placed accident-
ally upon the stem of an acanthus plant, which, shooting
forth, enclosed the basket with its foliage, some of which,
reaching the tile, turned gracefully over in the form of a
volute.
A celebrated sculptor, Calimachus, saw the basket thus
•decorated, and from the hint which it suggested conceived
and constructed a capital for a column. This was called
Corinthian, from the fact that it was invented and first made
use of at Corinth.
The Corinthian being the gayest, the richest, the most
lovely of all the orders, it is appropriate for edifices which
are dedicated to amusement, banqueting, and festivity — for
all places where delicacy, gavety, and splendor are desir-
able.
39. — Pcr§ian§ and Caryatides. — In addition to the three
regular orders of architecture, it was customary among the
Greeks and other nations to employ representations of the
human form, instead of columns, to support entablatures ;
these were called Persians and Caryatides.
40. — Persian* : are statues of men, and are so called in
commemoration of a victory gained over the Persians by
Pausanias. The Persian prisoners were brought to Athens
PROPORTIONS OF GRECIAN CORINTHIAN.
•x
sl
-
J
FIG. "j. — GRECIAN CORINTHIAN.
26 ARCHITECTURE.
and condemned to abject slavery ; and in order to represent .
them in the lowest state of servitude and degradation, the
statues were loaded with the heaviest entablature, the Doric.
41. — Caryatides: are statues of women dressed in long
robes after the Asiatic manner. Their origin is as follows :
In a war between the Greeks and the Caryans, the latter
were totally vanquished, their male population extinguished,
and their females carried to Athens. To perpetuate the
memory of this event, statues of females, having the form
and dress of the Caryans, were erected, and crowned with
the Ionic or Corinthian entablature. The caryatides were
generally formed of about the human size, but the persians
much larger, in order to produce the greater awe and
astonishment in the beholder. The entablatures were pro-
portioned to a statue in like manner as to a column of the
same height.
These semblances of slavery have been in frequent use
among moderns as well as ancients ; and, as a relief from
the stateliness and formality of the regular orders, are capa-
ble of forming a thousand varieties ; yet in a land of liberty
such marks of human degradation ought not to be perpetu-
ated.
42. — Roman Styles. — Strictly speaking, Rome had no
architecture of her own ; all she possessed was borrowed
from other nations. Before the Romans exchanged inter-
course with the Greeks, they possessed some edifices of
considerable extent and merit, which were erected by archi-
tects from Etruria ; but Rome was principally indebted to
Greece for what she acquired of the art. Although there is
no such thing as an architecture of Roman invention, yet
no nation, perhaps, ever was so devoted to the cultivation
of the art as the Roman. Whether we consider the number
and extent of their structures, or the lavish richness and
splendor with which they were adorned, we are compelled
to yield to them our admiration and praise. At one time,
under the consuls and emperors, Rome employed 400 ar-
chitects. The public works — such as theatres, circuses,
baths, aqueducts, etc. — were, in extent and grandeur, be-
PORTICO OF THE ERECTHEUM, ATHENS.
CHANGE OF STYLES BY THE ROMANS. 2?
yond anything- attempted in modern times. Aqueducts
were built to convey water from a distance of 60 miles or
more. In the prosecution of this work rocks and mountains
were tunnelled, and valleys bridged. Some of the latter
descended 200 feet below the level of the water; and. in
passing them the canals were supported by an arcade, or
succession of arches. Public baths are spoken of as large as
cities, being fitted up with numerous conveniences for ex-
ercise and amusement. Their decorations were most splen-
did ; indeed, the exuberance of the ornaments alone was
offensive to good taste. So overloaded with enrichments
were the baths of Diocletian that on one occasion of public
festivity great quantities of sculpture fell from the ceilings
and entablatures, killing many of the people.
43. — Grecian Order§ modified by the Romans. — The
orders of Greece were introduced into Rome in all their
perfection. But the luxurious Romans, not satisfied with
the simple elegance of their refined proportions, sought to
improve upon them by lavish displays of ornament. They
transformed in many instances the true elegance of the
Grecian art into a gaudy splendor, better suited to their
less refined taste. The Romans remodelled each of the
orders : the Doric (Fig. 8) was modified by increasing the
height of the column to 8 diameters ; by changing the
echinus of the capital for an ovolo, or quarter-round, and
adding an astragal and neck below it ; by placing the centre,
instead of one edge, of the first triglyph over the centre of
the column ; and introducing horizontal instead of inclined
mutules in the cornice, and in some instances dispensing
with them altogether. The Ionic was modified by diminish-
ing the size of the volutes, and, in some specimens, intro-
ducing a new capital in which the volutes were diagonally
arranged (Fig. 9). This new capital has been termed modern
Ionic. The favorite order at Rome and her colonies was
the Corinthian (Fig. 10). But this order the Roman artists,
in their search for novelty, subjected to many alterations —
especially in the foliage of its capital. Into the upper part
of this they introduced the modified Ionic capital ; thus
28
ARCHITECTURE.
combining the two in one. This change was dignified with
the importance of an order, and received the appellation
+ «
[29.
Vft
'
j^'*
n» >»<y
x>L->xL
FIG. 8.— ROMAN DORIC.
of COMPOSITE, or Roman : the best specimen of which is
found in the Arch of Titus (Fig. n). This style was not
PROPORTIONS OF THE ROMAN IONIC.
29
If4
.-._
aygpg^^
]^>34&*^&&1&1>S<1&1&Z>
wmm:
FIG, 9. — ROMAN IONIC.
30 ARCHITECTURE.
much used among the Romans themselves, and is but
slightly appreciated now.
44. — Tlie Tuscan Order: is said to have been intro-
duced to the Romans by the Etruscan architects, and to
have been the only style used in Italy before the introduc-
tion of the Grecian orders. However this may be, its simi-
larity to the Doric order gives strong indications of its
having been a rude imitation of that style : this is very prob-
able, since history informs us that the Etruscans held inter-
course with the Greeks at a remote period. The rudeness
of this order prevented its extensive use in Italy. All that
is known concerning it is from Vitruvius, no remains of
buildings in this style being found among ancient ruins.
For mills, factories, markets, barns, stables, etc., where
utility and strength are of more importance than beauty,
the improved modification of this order, called the modern
Tuscan (Fig. 12), will be useful; and its simplicity recom-
mends it where economy is desirable.
45. — Egyptian Style. — The architecture of the ancient
Egyptians — to which that of the ancient Hindoos bears
some resemblance — is characterized by boldness of outline,
solidity, and grandeur. The amazing labyrinths and exten-
sive artificial lakes, the splendid palaces and gloomy ceme-
teries, the gigantic pyramids and towering obelisks, of the
Egyptians were works of immensity and durability ; and
their extensive remains are enduring proofs of the enlight-
ened skill of this once-powerful but long since extinct na-
tion. The principal features of the Egyptian style of archi-
tecture are — uniformity of plan, never deviating from right
lines and angles ; thick walls, having the outer surface
slightly deviating inwardly from the perpendicular ; the
whole building low ; roof flat, composed of stones reaching
in one piece from pier to pier, these being supported by
enormous columns, very stout in proportion to their height ;
the shaft sometimes polygonal, having no base but with a
great variety of handsome capitals, the foliage of these being
of the palm, lotus, and other leaves ; entablatures having
simply an architrave, crowned with a huge cavetto orna-
PROPORTIONS OF THE ROMAN CORINTHIAN.
UNIVERSITY
FIG. io.— ROMAN CORINTHIAN.
ARCHITECTURE.
*»« WMWKM^-Q^WW&AWW^^
Fie. ii.— COMPOSITE ORDER— ARCH OF TITUS.
MASSIVENESS OF EGYPTIAN STRUCTURES. 33
mented with sculpture ; and the intercolumniation very nar-
row, usually i£ diameters and seldom exceeding 2^. In the
remains of a temple the walls were found to be 24 feet thick ;
and at the gates of Thebes, the walls at the foundation were
50 feet thick and perfectly solid. The immense stones of
which these, as well as Egyptian walls generally, were built,
had both their inside and outside surfaces faced, and the
oints throughout the body of the wall as perfectly close as
upon the outer surface. For this reason, as well as that the
buildings generally partake of the pyramidal form, arise
their great solidity and durability. The dimensions and ex-
tent of the buildings may be judged from the temple of
Jupiter at Thebes, which was 1400 feet long and 300 feet
wide — exclusive of the porticos, of which there was a great
number.
It is estimated by Mr. Gliddon, U. S. Consul in Egypt,
that not less than 25,000,000 tons of hewn stone were em-
ployed in the erection of the Pyramids of Memphis alone —
or enough to construct 3000 Bunker Hill monuments. Some
of the blocks are 40 feet long, and polished with emery to a
surprising degree. It is conjectured that the stone for these
pyramids was brought, by rafts and canals, from a distance
of six or seven hundred miles.
The general appearance of the Egyptian style of archi-
tecture is that of solemn grandeur — amounting sometimes to
sepulchral gloom. For this reason it is appropriate for cem-
eteries, prisons, etc. ; and being adopted for these purposes,
it is gradually gaining favor.
A great dissimilarity exists in the proportion, form, and
general features of Egyptian columns. In some instances,
there is no uniformity even in those of the same building,
each differing from the others either in its shaft or capital.
For practical use in this country, Fig. 13 may be taken as a
standard of this style. The Halls of Justice in Centre
Street, New York City, is a building in general accordance
with the principles of Egyptian architecture.
46. — Buildings in General. — In selecting a style for an
edifice, its peculiar requirements must be allowed to govern.
34
ARCHITECTURE.
736*
in
ir,
41
Fu;. 12. — MODIFIED TUSCAN ORDER.
FITNESS OF STYLES. 35
That style of architecture is to be preferred in which utility,
stability, and regularity are gracefully blended with gran-
deur and elegance. But as an arrangement designed for a
warm country would be inappropriate for a colder climate,
it would seem that the style of building ought to be modified
to suit the wants of the people for whom it is. designed.
High roofs to resist the pressure of heavy snows, and ar-
rangements for artificial heat, are indispensable in northern
climes ; while they would be regarded as entirely out of
place in buildings at the equator.
Among the Greeks, architecture was employed chiefly
upon their temples and other large buildings ; and the pro-
portions of the orders, as determined by them, when execu-
ted to such large dimensions, have the happiest effect. But
when used for small buildings, porticos, porches, etc., espe-
cially in country places, they are rather heavy and clumsy ;
in such cases, more slender proportions will be found to pro-
duce a better effect. The English cottage-style is rather
more appropriate, and is becoming extensively practised for
small buildings in the country.
47. — Expression. — Every building should manifest its
destination. If it be intended for national purposes, it
should be magnificent — grand ; for a private residence, neat
and modest ; for a banqueting-house, gay and splendid ; for
a monument or cemetery, gloomy — melancholy ; or, if for a
church, majestic and graceful — by some it has been said,
" somewhat dark and gloomy, as being favorable to a devo-
tional state of feeling ;" but such impressions can only re-
sult from a misapprehension of the nature of true devotion.
" Her ways are ways of pleasantness, and all her paths are
peace." The church should rather be a type of that
brighter world to which it leads. Simply for purposes of
contemplation, however, the glare of the noonday light
should be excluded, that the worshipper may, with Milton —
"Love the high, embowed roof,
With antique pillars massy pr£*f,
And storied windows richlyjj^ght,
Casting a dim, religious ligHt."
ARCHITECTURE.
H.P.
PREVALENCE OF WOODEN DWELLINGS. 37
However happily the several parts of an edifice may be
disposed, and however pleasing it may appear as a whole,
yet much depends upon its site, as also upon the character
and style of the structures in its immediate vicinity, and the
degree of cultivation of the adjacent country. A splendid
country-seat should have the out-houses and fences in the
same style with itself, the trees and shrubbery neatly
trimmed, and the grounds well cultivated.
48. — Durability. — Europeans express surprise that we
build so much with wood. And yet, in a new country,
where wood is plenty, that this should be so is no cause for
wonder. Still the practice should not be encouraged. Build-
ings erected with brick or stone are far preferable to those
of wood : they are more durable ; not so liable to injury by
fire, nor to need repairs ; and will be found in the end quite
as economical. A wooden house is suitable for a temporary
residence only ; and those who would bequeath a dwelling
to their children will endeavor to build with a more dura-
ble material. Wooden cornices and gutters, attached to
brick houses, are objectionable — not only on account of their
frail nature, but also because they render the building liable
to destruction by fire.
4-9. — Dwelling-Houses : are built of various dimensions
and styles, according to their destination ; and to give de-
signs and directions for their erection, it is necessary to know
their situation and object. A dwelling intended for a gar-
dener would require very different dimensions and arrange-
ments from one intended for a retired gentleman — with his
servants, horses, etc. ; nor would a house designed for the
city be appropriate for the country. For city houses, ar-
rangements that would be convenient for one family might
be very inconvenient for two or more. Figs. 14, 15, 16, 17,
1 8, and 19 represent the icJinograpJiical projection, or ground-
plan, of the floors of an ordinary city house, designed to be
occupied by one family only. Fig. 21 is an elevation, or
front view, of the same house. All these plans are drawn at
the same scale — which is that at the bottom of
38 ARCHITECTURE.
Fig. 14 is a Plan of the Under-Cellar.
a, is the coal-vault, 6 by 10 feet.
b, is the furnace for heating the house.
<:, d, are front and rear areas.
Fig. 15 is a Plan of the Basement.
a, is the library, or ordinary dining-room, 15 by 20 feet.
by is the kitchen, 15 by 22 feet.
c, is the store-room, 6 by 9 feet.
d, is the pantry, 4 by 7 feet.
e, is the china closet, 4 by 7 feet.
fj is the servants' water-closet.
g, is a closet.
//, is a closet with a dumb-waiter to the first story above.
i, is an ash closet under the front stoop.
j, is the kitchen-range.
k, is the sink for washing and drawing water.
/, are wash-trays.
Fig. 1 6 is a Plan of the First Story.
a, is the parlor, 1 5 by 34 feet.
b, is the dining-room, 16 by 23 feet.
c, is the vestibule.
<•, is the closet containing the dumb-waiter from the base-
ment.
/, is the closet containing butler's sink.
g, gy are closets.
//, is a closet for hats and cloaks.
iyjy are front and rear balconies.
Fig. 17 is the Second Story.
a, <?, are chambers, 15 by 13 feet.
b, is a bed-room, 7^ by 13 feet.
c, is the bath-room, 7^ by 13 feet.
d, dy are dressing-rooms, 6 by 7^ feet.
Cy e, are closets.
/, /, are wardrobes.
g, g, ve cupboards.
PLANS OF A CITY HOUSE.
39
FIG. 14.
UNDBR-CELLAR.
FIG- 15-
BASEMENT.
CITY DWELLING.
ARCHITECTURE.
CITY DWELLING.
FIG. 17.
SECOND STORY.
UPPER STORIES OF A CITY HOUSE.
Fig. 1 8 is the Third Story.
a, a, are chambers, 15 by 19 feet.
b, b, are bed-rooms, ;£ by 13 feet.
c, c, are closets.
d, is a linen-closet, 5 by 7 feet.
FIG. 19.
FOURTH STORY.
CITY DWELLING.
e, e, are dressing-closets.
f,f, are wardrobes.
g, g, are cupboards.
Fig. 19 is the Fourth Story.
a, a, are chambers, 14 by 17 feet.
42 ARCHITECTURE.
b, b, are bed-rooms, 8£ by 17 feet.
c, c, c, are closets.
d, is the step-ladder to the roof.
Fig. 20 is the Section of the House showing the heights
of the several stories.
Fig. 2 1 is the Front Elevation.
The size of the house is 25 feet front by 55 feet deep ; this
is about the average depth, although some are extended to
60 and 65 feet in depth.
These are introduced to give some general ideas of the
principles to be followed in designing city houses. In plac-
ing the chimneys in the parlors, set the chimney-breasts
equidistant from the ends of the room. The basement
chimney-breasts may be placed nearly in the middle of the
side of the room, as there is but one flue to pass through
the chimney-breast above ; but in the second story, as there
are two flues, one from the basement and one from the par-
lor, the breast will have to be placed nearly perpendicular
over the parlor breast, so as to receive the flues within the
jambs of the fire-place. As it is desirable to have the chim-
ney-breast as near the middle of the room as possible, it may
be placed a few inches towards that point from over the
breast below. So in arranging those of the stories above,
always make provision for the flues from below.
50. — Arranging the Stairs and Window§. — There should
be at least as much room in the passage at the side of the
stairs as upon them ; and in regard to the length of the pas-
sage in the second story, there must be room for the doors
which open from each of the principal rooms into the hall,
and more if the stairs require it. Having assigned a posi-
tion for the stairs of the second story, now generally placed
in the centre of the depth of the house, let the winders of
the other stories be placed perpendicularly over and under
them ; and be careful to provide for head-room. To ascer-
tain this, when it is doubtful, it is well to draw a vertical
section of the whole stairs ; but in ordinary cases this is not
FRONT OF A CITY HOUSE.
43
CITY DWELLING.
44 ARCHITECTURE.
necessary. To dispose the windows properly, the middle
window of each story should be exactly in the middle of the
front ; but the pier between the two windows which light
the parlor should be in the centre of that room ; because
when chandeliers or any similar ornaments hang from the
centre-pieces of the parlor ceilings, it is important, in order
to give the better effect, that the pier-glasses at the front
and rear be in a range with them. If both these objects
cannot be attained, an approximation to each must be at-
tempted. The piers should in no case be less in width than
the window openings, else the blinds or shutters, when
thrown open, will interfere with one another ; in general
practice, it is well to make the outside piers f of the width
of one of the middle piers. When this is desirable, deduct
the amount of the three openings from the width of the
front, and the remainder will be the amount of the width of
all the piers ; divide this by 10, and the product will be ^ of
a middle pier ; and then, if the parlor arrangements do not
interfere, give twice this amount to each corner pier, and
three times the same amount to each of the middle piers.
51. — Principles of Architecture. — To build well requires
close attention and much experience. The science of build-
ing is the result of centuries of study. Its progress towards
perfection must have been exceedingly slow. In the con-
struction of the first frail and rude habitations of men, the
primary object was, doubtless, utility — a mere shelter from
sun and rain. But as successive storms shattered his poor
tenement, man was taught by experience the necessity of
building with an idea to durability. And as the symmetry,
proportion, and beauty of nature met his admiring gaze,
contrasting so strangely with the misshapen and dispropor-
tioned work of his own hands, he was led to make gradual
changes, till his abode was rendered not only commodious
and durable, but pleasant in its appearance ; and building
became a fine art, having utility for its basis.
52. — Arrangement. — In all designs for buildings of im-
portance, utility, durability, and beauty, the first great prin-
ciples, should be pre-eminent. In order that the edifice be
• Ox
UNIVERSITY
ESSENTIAL REQUIREMENTS OF A BUILDING. 45
useful, commodious, and comfortable, the arrangement of the
apartments should be such as to fit them for their several
destinations ; for publio assemblies, oratory, state, visitors,
retiring, eating, reading, sleeping, bathing, dressing, etc.—
these should each have its own peculiar form and situation.
To accomplish this, and at the same time to make their
relative situation agreeable and pleasant, producing regu-
larity and harmony, require 'in some instances much skill
and sound judgment. Convenience and regularity are very
important, and each should have due attention ; yet when
both cannot be obtained, the latter should in most cases
give place to the former. A building that is neither con-
venient nor regular, whatever other good qualities it may
possess, will be sure of disapprobation.
53. — Ventilation. — Attention should be given to such
arrangements as are calculated to promote health : among
these, ventilation is by no means the least. For this pur-
pose, the ceilings of the apartments should have a respect-
able height ; and the sky-light, or any part of the roof that
can be made movable, should be arranged with cord and
pullies, so as to be easily raised and lowered. Small open-
ings near the ceiling, that may be closed at pleasure, should
be made in the partitions that separate the rooms from the
passages — especially for those rooms which are used for
sleeping apartments. All the apartments should be so ar-
ranged as to secure their being easily kept dry and clean.
In dwellings, suitable apartments should be fitted up for
bathing with all the necessary apparatus for conveying
water.
54. — Stability. — To secure this, an edifice should be de-
signed upon well-known geometrical principles : such as
science has demonstrated to be necessary and sufficient for
firmness and durability. It is well, also, that it have the
appearance of stability as well as the reality ; for should it
seem tottering and unsafe, the sensation of fear, rather than
those of admiration and pleasure, Avill be excited in the be-
holder. To secure certainty and accuracy in the applica-
tion of those principles, a knowledge of the strength and
46 ARCHITECTURE.
other properties of the materials used is indispensable ; and
in order that the whole design be so made as to be capable
of execution, a practical knowledge of the requisite mechan-
ical operations is quite important.
55. — Decoration. — The elegance of a design, although
chiefly depending upon a just proportion and harmony of
the parts, will be promoted by the introduction of orna-
ments, provided this be judiciously performed ; for enrich-
ments should not only be ot a proper character to suit the
style of the building, but should also have their true posi-
tion, and be bestowed in proper quantity. The most com-
mon fault, and one which is prominent in Roman architec-
ture, is an excess of enrichment : an error which is carefully
to be guarded against. But those who take the Grecian
models for their standard will not be liable to go to that
extreme. In ornamenting a cornice, or any other assem-
blage ol mouldings, at least every alternate member should
be left plain ; and those that are near the eye should be more
finished than those which are distant. Although the charac-
teristics of good architecture are utility and elegance, in
connection with durability, yet some buildings are designed
expressly for use, and others again for ornament : in the
former, utility, and in the latter, beauty, should be the gov-
erning principle.
56. — Elementary Parts of a Building. — The builder
should be acquainted with the principles upon which the
essential, elementary parts of a building are founded. A
scientific knowledge of these will insure certainty and secu-
rity, and enable the mechanic to erect the most extensive
and lofty edifices with confidence. The more important
parts are the foundation, the column, the wall, the lintel,
the arch, the vault, the dome, and the roof. A separate
description of the peculiarities of each would seem to be
necessary, and cannot perhaps be better expressed than in
the following language of a modern writer on this subject,
slightly modified :
UNIVERSITY
STRUCTURAL FEATURES OF A BUILDING. 47
57. — The Foundation: of a building should be begun
at a certain depth in the earth, to secure a solid basis, below
the reach of frost and common accidents. The most solid
basis is rock, or gravel which has not been moved. Next
to these are clay and sand, provided no other excavations
have been made in the immediate neighborhood. From
this basis a stone wall is carried up to the surface of the
ground, and constitutes the foundation. Where it is in-
tended that the superstructure shall press unequally, as at
its piers, chimneys, or columns, it is sometimes of use to
occupy the space between the points of pressure by an
inverted arch. This distributes the pressure equally, and
prevents the foundation from springing between the differ-
ent points. In loose or muddy situations, it is always un-
safe to build, unless we can reach the solid bottom below.
In salt marshes and flats, this is done by depositing timbers,
or driving wooden piles into the earth, and raising walls
upon them. The preservative quality of the salt will keep
these timbers unimpaired for a great length of time, and
makes the foundation equally secure with one of brick or
stone.
58. — The Column, or Pillar: is the simplest member in
any building, though by no means an essential one to all.
This is a perpendicular part, commonly of equal breadth
and thickness, not intended for the purpose of enclosure,
but simply for the support of some part of the superstruc-
ture. The principal force which a column has to resist is
that of perpendicular pressure. In its shape, the shaft of a
column should not be exactly cylindrical, but, since the
lower part must support the weight of the superior part, in
addition to the weight which presses equally on the whole
column, the thickness should gradually decrease from bot-
tom to top. The outline of columns should be a little
curved, so as to represent a portion of a very long spheroid,
or paraboloid, rather than of a cone. This figure is the joint
result of two calculations, independent of beauty of appear-
ance. One of these is, that the form best adapted for sta-
bility of base is that of a cone ; the other is, that the figure,
48 ARCHITECTURE.
which would be of equal strength throughout for support-
ing a superincumbent weight, would be generated by the
revolution of two parabolas round the axis of the column,
the vertices of the curves being at its extremities. The
swell of the shafts of columns was called the entasis by the
ancients. It has been lately found that the columns of the
Parthenon, at Athens, which have been commonly supposed
straight, deviate about an inch from a straight line, and that
their greatest swell is at about one third of their height.
Columns in the antique orders are usually made to diminish
one sixth or one seventh of their diameter, and sometimes
even one fourth. The Gothic pillar is commonly of equal
thickness throughout.
59. — The Wall : another elementary part of a building,
may be considered as the lateral continuation of the column,
answering the purpose both of enclosure and support. A
wall must diminish as it rises, for the same reasons, and in
the same proportion, as the column. It must diminish still
more rapidly if it extends through several stories, support-
ing weights at different heights. A wall, to possess the
greatest strength, must also consist of pieces, the upper and
lower surfaces of which are horizontal and regular, not
rounded nor oblique. The walls of most of the ancient
structures which have stood to the present time are con-
structed in this manner, and frequently have their stones
bound together with bolts and clamps of iron. The same
method is adopted in such modern structures as are intended
to possess great strength and durability, and, in some cases,
the stones are even dovetailed together, as in the light-
houses at Eddy stone and Bell Rock. But many of our
modern stone walls, for the sake of cheapness, have only one
face of the stones squared, the inner half of the wall being
completed with brick ; so that they can, in reality, be con-
sidered only as brick walls faced with stone. Such walls are
said to be liable to become convex outwardly, from the dif-
ference in the shrinking of the cement. Rubble walls are
made of rough, irregular stones, laid in mortar. The stones
should be broken, if possible, so as to produce horizontal
VARIOUS METHODS OF ERECTING WALLS. 49
surfaces. The coffer walls of the ancient Romans were made
by enclosing successive portions of the intended wall in a
box, and filling it with stones, sand, and mortar promis-
cuously. This kind of structure must have been extremely
insecure. The Pantheon and various other Roman build-
ings are surrounded with a double brick wall, having its
vacancy filled up with loose bricks and cement. The whole
has gradually consolidated into a mass of great firmness.
60. — The Reticulated Walls : of the Romans — com-
posed of bricks with oblique surfaces — would, at the present
day, be thought highly unphilosophical. Indeed, they could
not long have stood, had it not been for the great strength
of their cement. Modern brick walls are laid with great
precision, and depend for firmness more upon their position
than upon the strength of their cement. The bricks being
laid in horizontal courses, and continually overlaying each
other, or breaking joints, the whole mass is strongly inter-
woven, and bound together. Wooden walls, composed of
timbers covered with boards, are a common but more per-
ishable kind. They require to be constantly covered with a
coating of a foreign substance, as paint or plaster, to pre-
serve them from spontaneous decomposition. In some parts
of France, and elsewhere, a kind of wall is made of earth,
rendered compact by ramming it in moulds or cases. This
method is called building in pise, and is much more durable
than the nature of the material would lead us to suppose.
Walls of all kinds are greatly strengthened by angles and
curves, also by projections, such as pilasters, chimneys, and
buttresses. These projections serve to increase the breadth
of the foundation, and are always to be made use of in large
buildings, and in walls of considerable length.
61. — The Lintel, or Beam: extends in a right line over
a vacant space, from one column or wall to another. The
strength of the lintel will be greater in proportion as its
transverse vertical diameter exceeds the horizontal, the
strength being always as the square of the depth. The
floor is the lateral continuation or connection of beams by
means of a covering of boards.
50 ARCHITECTURE.
62.— The Arch : is a transverse member of a building,
answering the same purpose as the lintel, but vastly exceed-
ing it in strength. The arch, unlike the lintel, may consist
of any number of constituent pieces, without impairing its
strength. It is, however, necessary that all the pieces should
possess a uniform shape, — the shape of a portion of a
wedge, — and that the joints, formed by the contact of their
surfaces, should point towards a common centre. In this
case, no one portion of the arch can be displaced or forced
inward ; and the arch cannot be broken by any force which
is not sufficient to crush the materials of which it is made.
In arches made of common bricks, the sides of which are
parallel, any one of the bricks might be forced inward, were
it not for the adhesion of the cement. Any two of the bricks,
however, by the disposition of their mortar, cannot collect-
ively be forced inward. An arch of the proper form, when
complete, is rendered stronger, instead of weaker, by the
pressure of a considerable weight, provided this pressure be
uniform. While building, however, it requires to be sup-
ported by a centring of the shape of its internal surface,
until it is complete. The upper stone of an arch is called
the keystone, but is not more essential than any other. In
regard to the shape of the arch, its most simple form is that
of the semicircle. It is, however, very frequently a smaller
arc of a circle, or a portion of an ellipse.
63. — Hooke'§ Theory of an Arch. — The simplest theory
of an arch supporting itself only is that of Dr. Hooke.
The arch, when it has only its own weight to bear, may be
considered as the inversion of a chain, suspended at each
end. The chain hangs in such a form that the weight of
each link or portion is held in equilibrium by the result of
two forces acting at its extremities ; and these forces, or
tensions, are produced, the one by the weight of the portion
of the chain below the link, the other by the same weight
increased by that of the link itself, both of them acting ori-
ginally in a vertical direction. Now, supposing the chain
inverted, so as to constitute an arch of the same form and
weight, the relative situations of the forces will be the same,
VIADUCT AT CHAUMONT.
PECULIARITIES OF THE ARCH. 51
only they will act in contrary directions, so that they are
compounded in a similar manner, and balance each other on
the same conditions.
The arch thus formed is denominated a catenary arch.
In common cases, it differs but little from a circular arch of
the extent of about one third of a whole circle, and rising
from the abutments with an obliquity of about 30 degrees
from a perpendicular. But though the catenary arch is the
best form for supporting its own weight, and also all addi-
tional weight which presses in a vertical direction, it is not
the best form to resist lateral pressure, or pressure like that
of fluids, acting equally in all directions. Thus the arches
of bridges and similar structures, when covered with loose
stones and earth, are pressed sideways, as well as vertically,
in the same manner as if they supported a weight of fluid.
In this case, it is necessary that the arch should arise more
perpendicularly from the abutment, and that its general fig-
ure should be that of the longitudinal segment of an ellipse.
In small arches, in common buildings, where the disturbing
force is not great, it is of little consequence what is the
shape of the curve. The outlines may even be perfectly
straight, as in the tier of bricks which we frequently see
over a window. This is, strictly speaking, a real arch, pro-
vided the surfaces of the bricks tend toward a common
centre. It is the weakest kind of arch, and a part of it
is necessarily superfluous, since no greater portion can act
in supporting a weight above it than can be included be-
tween two curved or arched lines.
f
64. — Gothic Arches. — Besides these arches, various
others are in use. The acute or lancet arch, much used in
Gothic architecture, is described usually from two cen-
tres outside the arch. It is a strong arch for supporting
vertical pressure. The rampant arch is one in which the two
ends spring from unequal heights. The Jwrseshoe or Moorisli
arch is described from one or more 'centres placed above the
base line. In this arch, the lower parts are in danger of
being forced inward. The ogee arch is concavo-convex, and
therefore fit only for ornament.
52 ARCHITECTURE.
65. — Arch: Definition*; Principles. — The upper sur-
face is called the extrados, and the inner, the intrados.
The spring is where the intrados meets the abutments. The
span is the distance between the abutments. The wedge-
shaped stones which form an arch are sometimes called
vonssoirs, the uppermost being- the keystone. The part of a
pier from which an arch springs is called the impost, and the
curve formed by the under side of the voussoirs, the archi-
volt. It is necessary that the walls, abutments, and piers on
which arches are supported should be so firm as to resist the
lateral thrust, as well as vertical pressure, of the arch.
It will at once be seen that the lateral or sideway pressure
of an arch is very considerable, when we recollect that every
stone, or portion of the arch, is a wedge, a part of whose
force acts to separate the abutments. For want of attention
to this circumstance, important mistakes have been commit-
ted, the strength of buildings materially impaired, and their
ruin accelerated. In some cases, the want of lateral firmness
in the walls is compensated by a bar of iron stretched across
the span of the arch, and connecting the abutments, like the
tie-beam of a roof. This is the case in the cathedral of Milan
and some other Gothic buildings.
66. — An Arcade : or continuation of arches, needs only
that the outer supports of the terminal arches should be
strong enough to resist horizontal pressure. In the inter-
mediate arches, the lateral force of each arch is counter-
acted by the opposing lateral force of the one contiguous to
it. In bridges, however, where individual arches are liable
to be destroyed by accident, it is desirable that each of the
piers should possess sufficient horizontal strength to resist
the lateral pressure of the adjoining arches.
67. — The Vault: is the lateral continuation of an arch,
serving to cover an area or passage, and bearing the same
relation to the arch that the wall does to the column. A
simple vault is constructed on the principles of the arch, and
distributes its pressure equally along the walls or abutments.
A complex or groined vault is made by two vaults intersect^
ing each other, in which case the pressure is thrown upon
VARIOUS CONSTRUCTIONS OF THE DOME. 53
springing points, and is greatly increased at those points.
The groined vault is common in Gothic architecture.
68. — The Dome: sometimes called cupola,'^ a concave
covering to a building, or part of it, and may be either a
segment of a sphere, of a spheroid, or of any similar figure.
When built of stone, it is a very strong kind of structure,
even more so than the arch, since the tendency of each part
to fall is counteracted, not only by those above and below it,
but also by those on each side. It is only necessary that
the constituent pieces should have a common form, and that
this form should be somewhat like the frustum of a pyra-
mid, so that, when placed in its situation, its four angles may
point toward the centre, or axis, of the dome. During the
erection of a dome, it is not necessary that it should be sup-
ported by a centring, until complete, as is done in the arch.
Each circle of stones, when laid, is capable of supporting
itself without aid from those above it. It follows that the
dome may be left open at top, without a keystone, and
yet be perfectly secure in this respect, being the reverse of
the arch. The dome of the Pantheon, at Rome, has been
always open at top, and yet has stood unimpaired for nearly
2006 years. The upper circle of stones, though apparently
the weakest, is nevertheless often made to support the addi-
tional weight of a lantern or tower above it. In several of
the largest cathedrals, there are two domes, one within the
other, which contribute their joint support to the lantern,
which rests upon the top. In these buildings, the dome
rests upon a circular wall, which is supported, in its turn, by
arches upon massive pillars or piers. This construction is
called building upon pendentivcs, and gives open space and
room for passage beneath the dome.' The remarks which
have been made in regard to the abutments of the arch
apply equally to the walls immediately supporting a dome.
They must be of sufficient thickness and solidity to resist
the lateral pressure of the dome, which is very great. The
walls of the Roman Pantheon are of great depth and solid-
ity. In order that a dome in itself should be perfectly
secure, its lower parts must not be too nearly vertical, since,
54 ARCHITECTURE.
in this case, they- partake of the nature of perpendicular
walls, and are acted upon by the spreading force of the
parts above them. The dome of St. Paul's Church, in Lon-
don, and some others of similar construction, are bound with
chains or hoops of iron, to prevent them from spreading at
bottom. Domes which are made of wood depend, in part*
for their strength on their internal carpentry. The Halle
du Bled, in Paris, had originally a wooden dome more than
200 feet in diameter, and only one foot in thickness. This
has since been replaced by a dome of iron. (See Art-
235.)
69. — Tlie Roof: is the most common and cheap method
of covering buildings, to protect them from rain and other
effects of the weather. It is sometimes flat, but more fre-
quently oblique, in its shape. The flat or platform roof is
the least advantageous for shedding rain, and is seldom used
in northern countries. The pent roof, consisting of two
oblique sides meeting at top, is the most common form.
These roofs are made steepest in cold climates, where they
are liable to be loaded with snow. Where the four sides of
the roof are all oblique, it is denominated a hipped roof, and
where there are two portions to the roof, of different ob-
liquity, it is a curb, or mansard roof. In modern times, roofs
are made almost exclusively of wood, though frequently
covered with incombustible materials. The internal struc-
ture or carpentry of roofs is a subject of considerable me-
chanical contrivance. The roof is supported by rafters,
which abut on the walls on each side, like the extremities of
an arch. If no other timbers existed except the rafters,
they would exert a strong lateral pressure on the walls,
tending to separate and overthrow them. To counteract
this lateral force, a tic-beam, as it is called, extends across,
receiving the ends of the rafters, and protecting the wall
from their horizontal thrust. To prevent the tie-beam from
sagging, or bending downward with its own weight, a king-
post is erected from this beam, to the upper angle of the
rafters, serving to connect the whole, and to suspend the
weight of the beam. This is called trussing. Queen-posts
MANNER OF CONSTRUCTING ROOFS. 55
are sometimes added, parallel to the king-post, in large roofs I
also various other connecting timbers. In Gothic buildings,
where the vaults do not admit of the use of a tie-beam, the
rafters are prevented from spreading, as in an arch, by the
strength of the buttresses.
In comparing the lateral pressure of a high roof with
that of a low one, the length of the tie-beam being the same,
it will be seen that a high roof, from its containing most
materials, may produce the greatest pressure, as far as
weight is concerned. On the other hand, if the weight of
both be equal, then the low roof will exert the greater press-
ure ; and this will increase in proportion to the distance of
the point at which perpendiculars, drawn from the end of
each rafter, would meet. In roofs, as well as in wooden
domes and bridges, the materials are subjected to an inter-
nal strain, to resist which the cohesive strength of the ma-
terial is relied on. On this account, beams should, when
possible, be of one piece. Where this cannot be effected,
two or more beams are connected together by splicing.
Spliced beams are never so strong as whole ones, yet they
may be made to approach the same strength, by affixing lat-
eral pieces, or by making the ends overlay each other, and
connecting them with bolts and straps of iron. The ten-
dency to separate is also resisted, by letting the two pieces
into each other by the process called scarfing. Mortices, in-
tended to truss or suspend one piece by another, should be
formed upon similar principles.
Roofs in the United States, after being boarded, receive
a secondary covering of shingles. When intended to be
incombustible, they are covered with slates or earthen tiles,
or with sheets of lead, copper, or tinned iron. Slates are
preferable to tiles, being lighter, and absorbing less moisture.
Metallic sheets are chiefly used for flat roofs, wooden domes,
and curved and angular surfaces, which require a flexible
material to cover them, or have not a sufficient pitch to shed
the rain from slates or shingles. Various artificial composi-
tions are occasionally used to cover roofs, the most common
of which are mixtures of tar with lime, and sometimes with
sand and gravel. — Ency. Am. (See Art. 202.)
SECTION II.— CONSTRUCTION.
ART. 70. — Con§tructioii E§seiUiai. — Construction is that
part of the Science of Building which treats of the Laws of
Pressure and the strength of materials. To the architect
and builder a knowledge of it is absolutely essential. It de-
serves a larger place in a volume of this kind than is gene-
rally allotted to it. Something, indeed, has been said upon
the styles and principles, by which the best arrangements
may be ascertained ; yet, besides this, there is much to be
learned. For however precise or workmanlike the several
parts may be made, what will it avail, should the system of
framing, from deficient material, or an erroneous position of
its timbers, fail to sustain even its own weight ? Hence the
necessity for a knowledge of the laws of pressure and the
strength of materials. These being once understood, we
can with confidence determine the best position and dimen-
sions for the several pieces which compose a floor or a roof,
a partition or a bridge. As systems of framing are more or
less exposed to heavy weights and strains, and, in case of
failure, cause not only a loss of labor and material, but fre-
quently that of life itself, it is vety important that the mate-
rials employed be of the proper quantity and quality to serve
their destination. And, on the other hand, any superfluous
material is not only useless, but a positive injury, as it is an
unnecessary load upon the points of support. It is neces-
sary, therefore, to know the least quantity of material that
will suffice for strength. Not the least common fault in
framing is that of using an excess of material. Economy, at
least, would seem to require that this evil be abated. <
Before proceeding to consider the principles upon which
a system of framing should be constructed, let us attend to
a few of the elementary laws in Mechanics, which will be
found to be of great value in determining those principles.
INTERIOR OF THE CATHEDRAL, SIENNA.
DIRECT AND OBLIQUE SUPPORTS.
57
71. — ILaws of Pre§§ure. — (i.) A heavy body always ex-
erts a pressure equal to its own weight in a vertical direc-
tion. Example : Suppose an iron ball weighing 100 Ibs. be
supported upon the top of a perpendicular post (Fig. 22-A) ;
then the pressure exerted upon that post will be equal to
the weight of the ball, viz., 100 Ibs. (2.) But if two inclined
posts (Fig. 22-B) be substituted for the perpendicular sup-
port, the united pressures upon these posts will be more
than equal to the weight, and will be in proportion to their
position. The farther apart their feet are spread the greater
will be the pressure, and vice versa. Hence tremendous
strains may be exerted by a comparatively small weight.
And it follows, therefore, that a piece of timber intended
for a strut or post should be so placed that its axis may
coincide, as nearly as possible, with the direction of the
pressure. The direction of the pressure of the weight W
(Fig. 22-B) is in the vertical line b d\ and the weight W
would fall in that line if the two posts were removed ; hence
the best position for a support for the weight would be in
A.
FlG. 22.
that line. But as it rarely occurs in systems of framing
that weights can be supported by any single resistance, they
requiring generally two or more supports (as in the case of
a roof supported by its rafters), it becomes important, there-
fore, to know the exact amount of pressure any certain
weight is capable of exerting upon oblique supports. Now,
it has been ascertained that the three lines of a triangle,
drawn parallel with the direction of three concurring forces
in equilibrium, are in proportion respectively to these
58 CONSTRUCTION.
forces. For example, in Fig. 22-B, we have a representation
of three forces concurring in a point, which forces are in
equilibrium and at rest ; thus, the weight W is one force,
and the resistances exerted by the two pieces of timber are
the other two forces. The direction in which the first force
acts is vertical — downwards ; the direction of the two other
forces is in the axis of each piece of timber respectively.
These three forces all tend towards the point b.
Draw the axes a b and b c of the two supports ; make b d
vertical, and from d draw d e and d f parallel with the axes
b c and b a repectively. Then the triangle b d e has its
lines parallel respectively with the direction of the three
forces ; thus, bd is in the direction of the weight W, d e paral-
lel with the axis of the timber D, and e b is in the direction of
the timber C. In accordance with the principle above stated,
the lengths of the sides of the triangle b d e are in propor-
tion respectively to the three forces aforesaid ; thus —
As the length of the line b d
Is to the number of pounds in the weight W,
So is the length of the line £ e
To the number of pounds' pressure resisted by the
timber C.
Again —
As the length of the line b d
Is to the number of pounds in the weight W,
So is the length of the line d c
To the number of pounds' pressure resisted by the
timber D.
And again —
As the length of the line b e
Is to the pounds' pressure resisted by C,
So is the length of the line d e
To the pounds' pressure resisted by D.
These proportions are more briefly stated thus —
\st. bd\ W\\be\P,
P being used as a symbol to represent the number of pounds'
pressure resisted by the timber C.
PARALLELOGRAM OF FORCES. 59
2d. b d : W : : d e : Q,
Q representing the number of pounds' pressure resisted by
the timber D.
3d. b e \P\\de\ Q.
72. — Parallelogram of Forces. — This relation between
lines and pressures is applicable in ascertaining the pres-
sures induced by known weights throughout any system of
framing. The parallelogram b e df is called the Parallelo-
%ram of Forces ; the two lines If e- and bf being called the
components, and the line b d the resultant. Where it is re-
quired to find the components from a given resultant (Fig.
22- B\ the fourth line df/need not be drawn, for the triangle
b d e gives the desired result. But when the resultant is to
be ascertained from given components (Fig. 28), it is more
convenient to draw the fourth line.
73. — The Resolution of Forces : Is the finding of two
or more forces which, acting in different directions, shall
exactly balance the pressure of any given single force. To
make a practical application of this, let it be required to
ascertain the oblique pressure in Fig. 22-B. In this figure the
line bd measures half an inch (0-5 inch), and the line be
three tenths of an inch (0-3 inch). Now if the weight W
be supposed to be 1200 pounds, then the first stated propor-
tion above,
b d\ W: : b e : P, becomes 0-5 : 1200 : : 0-3 : P.
And since the product of the means divided by one of the
extremes gives the other extreme, this proportion may be
put in the form of an equation^ thus —
I200X 0-3 _p
0-5
Performing the arithmetical operation here indicated— -that
is, multiplying together the two quantities above the line,
and dividing the product by the quantity under the line— the
6o
CONSTRUCTION.
quotient will be equal to the quantity represented by P, viz.,
the pressure resisted by the timber C. Thus —
1 200
0-5)360-0
720 = P.
The strain upon the timber C is, therefore, equal to 720
pounds; and since, in this case (the two timbers being in-
clined equally from the vertical), the line e d is equal to the
line b e, therefore the strain upon the other timber D is
also 720 pounds.
FIG. 23.
74. — Inclination of Supports Unequal. — In Fig. 23 the
pressures in the two supports are unequal. The supports
are also unequal in length. The length of the supports,
however, does not alter the amount of pressure from the
concentrated load supported ; but generally long timbers
are not so capable of resistance as shorter ones. For, not
being so stiff, they bend more readily, and, since the com-
pression is in proportion to the length, they therefore
shorten more. To ascertain the pressures in Fig. 23, let the
weight suspended from b d be equal to two and three quarter
tons (2-75 tons). The line b d measures five and a half
tenths of an inch (0-55 inch), and the line b e half an inch
(0-5 inch). Therefore, the proportion
b d\ W \\ b e : P becomes 0-55 : 2-75 : : 0-5 : P,
STRAIN IN PROPORTION TO INCLINATION. — £r
and 2.75x0.5 /
0-55 (UNIVERSITY
0'55)i -375(2- 5 = ^.
I 10
275
275
The strain upon the timber A. is, therefore, equal to two
and a half tons.
Again, the line e d measures four tenths of an inch (0-4
inch) ; therefore, the proportion
b d \ W \ : e d \ Q becomes 0-55 : 2-75 : : 0-4 : Q,
and 2./5 x 0-4 __ ^
o-55
2-75
0-4
0-55) 1-100(2 = Q.
I 10
The strain upon the timber B is, therefore, equal to two
tons.
'/5. — The Strains Exceed tBie Weights. — Thus the united
pressures upon the two inclined supports always exceed the
weight. In the last case, 2f tons exert a pressure of 2\ and
2 tons, equal together to 4^ tons ; and in the former case,
1200 pounds exert a pressure of twice 720 pounds, equal to
1440 pounds. The smaller the angle of inclination to the '
horizontal, the greater will be the pressure upon the sup-
ports. So, in the frame of a roof, the strain upon the rafters
decreases gradually with the increase of the angle of incli-
nation to the horizon, the length of the rafter remaining the
same.
62 CONSTRUCTION.
This is true in comparing one system of framing with
another ; but in a system where the concentrated weight to
be supported is not in the middle (see Fig. 23), and, in con-
sequence, the supports are not inclined equally, the strain
will be greatest upon that support which has the greatest
inclination to the horizon.
76. — Minimum Thrust of Rafter§. — Ordinarily, as in
roofs, the load is not concentrated, it being that of the fram-
ing itself. Here the amount of the load will be in proportion
to the length of the rafter, and this will increase with the
increase of the angle of inclination, the span remaining the
same. So it is seen that in enlarging the angle of inclina-
tion to the horizon in order to lessen the oblique thrust, the
load is increased in consequence of the elongation of the
rafter, thus increasing the oblique thrust. Hence there is
an economical angle of inclination. A rafter will have the
least oblique thrust when its angle of inclination .to the
horizon is 35° 16' nearly. This angle is attained very nearly
when the rafter rises 8J inches per foot, or when the height
B C (Fig. 32) is to the base A C as 8£ is to 12, or as 0-7071 is
to i-o.
77. — Practical Method of Determining Strain§. — A com-
parison of pressures in timbers, according to their position,
may be readily made by drawing various designs of framing
and estimating the several strains in accordance with the
parallelogram of forces, always drawing the triangle b d e
so that the three lines shall be parallel with the three forces
or pressures respectively. The length of the lines forming
this triangle is unimportant, but it will be found more con-
venient if the line drawn parallel with the known force is
made to contain as many inches as the known force contains
pounds, or as many tenths of an inch as pounds, or as many
inches as to is, or tenths of an inch as tons; or, in general,
as many divisions of any convenient scale as there are units
of weight or pressure in the known force. If drawn in this
manner, then the number of divisions of the same scale
found in the other two lines of the triangle will equal the
units of pressure or weight of the other two forces respect-
HORIZONTAL THRUST OF RAFTERS. 63
ively, and the pressures sought will be ascertained simply by
applying the scale to the lines of the triangle.
For example, in Fig. 23, the vertical line b d, of the tri-
angle, measures fifty-five hundredths of an inch (0-55 inch) ;
the line b e, fifty hundredths (0-50 inch); and the line e d>
forty (0-40 inch). Now, if it be supposed that the vertical
pressure, or the weight suspended below b d, is equal to 55
pounds, then the pressure on A will equal 50 pounds, and
that on B will equal 40 pounds ; for, by the proportion above
stated,
b d: W-.-.b e:P,
55 : 55 :: 50: 50;
and so of the other pressure.
If a scale cannot be had of equal proportions with the
forces, the arithmetical process will be shortened somewhat
by making the line of the triangle that represents the known
weight equal to unity of a decimally divided scale, then the
other lines will be measured in tenths or hundredths ; and
in the numerical statement of the proportions between the
lines and forces, the first term being unity, the fourth term
will be ascertained simply by multiplying the second and
third terms together.
For example, if the three lines are i, 0-7, and 1-3, and
the known weight is 6 tons, then
b d : W : : b e : P becomes
I : 6 : : 0-7 : P — 4-2,
equals four and two tenths tons. Again —
bd\ W : : e d : Q becomes
I :6:: 1-3: Q = 7-8,
equals seven and eight tenths tons.
78.— Horizontal Thrust.— In Fig. 24, the weight ^presses
the struts in the direction of their length ; their feet, n n,
therefore, tend to move in the direction n o, and would so
move were they not opposed by a sufficient resistance from
the blocks, A and A. If a piece of each block be cut off at
64 CONSTRUCTION.
the horizontal line, a n, the feet of the struts would slide
away from each other along that line, in the direction na;
but if, instead of these, two pieces were cut off at the verti-
cal line, nb, then the struts would descend vertically. To
estimate the horizontal and the vertical pressures exerted by
the struts, let n o be made equal (upon any scale of equal
parts) to the number of tons with which the strut is pressed ;
FIG. 24.
construct the parallelogram of forces by drawing o c parallel
to an, and of parallel to bn; then nf (by the same scale)
shows the number of tons pressure that is exerted by the
strut in the direction n a, and n e shows the amount exerted
in the direction n b. By constructing designs similar to this,
giving various and dissimilar positions to the struts, and
then estimating the pressures, it will be found in every case
that the horizontal pressure of one strut is exactly equal to
that of the other, however much one strut may be inclined
more than the other; and also, that the united vertical
pressure of the two struts is exactly equal to the weight W.
(In this calculation the weight of the timbers has not been
taken into consideration, simply to avoid complication to
the learner. In practice it is requisite to include the weight
of the framing with the load upon the framing.)
Suppose that the two1 struts, B and B (Fig. 24), were
rafters of a roof, and that instead of the blocks, A and A, the
walls of a building were the supports: then, to prevent
TIES DESIRABLE IN ROOFS. 65
the walls from being thrown over by the thrust of B and B,
it would be desirable to remove the horizontal pressure.
Tnis may be done by uniting- the feet of the rafters with a
u
FIG. 25.
rope, iron rod, or piece of timber, as in Fig. 25. This figure
is similar to the truss of a roof. The horizontal strains on
the tie-beam, tending to pull it asunder in the direction of
its length, may be measured at the foot of the rafter, as was
shown at Fig* 24 ; but, it can be more readily and as accu-
rately measured by drawing from f and e horizontal lines to
the vertical line, b d, meeting it in o and o ; then/0 will be
the horizontal thrust at B, and co at A ; these will be found
to equal one another. When the rafters of a roof are thus
connected, all tendency to thrust out the walls horizontally
is removed, the only pressure on them is in a vertical direc-
tion, being equal to the weight of the roof and whatever it
has to support. This pressure is beneficial rather than
otherwise, as a roof having trusses thus formed, and the
trusses well braced to each other, tends to steady the walls.
79.— Position of Supports.—/^. 26 and 27 exhibit two
methods of supporting the equal weights, W and W. Let
it be required to measure and compare the strains produced
on the pieces, A B and A C. Construct the parallelogram of
forces, ebfd, according to Art. 71. Then 3/will show the
66
CONSTRUCTION.
strain on A B, and b e the strain on A C. By comparing the
figures, bd being equal in each, it will be seen that the
strains in Fig. 26 are about three times as great as those in
B
FIG. 27.
Fig. 27 ; the position of the pieces, A B and A C, in Fig. 27,
is therefore far preferable.
FIG. 28.
80.— The Composition of Force§ : consists in ascertain-
ing the direction and amount of one force which shall be
just capable of balancing tivo or more given forces, acting in
different directions. This is only the reverse of the resolu-
STRAINS INVOLVED IN CRANE.
tion of forces ; and the two are founded on one and the
same principle, and may be solved in the same manner. For
example, let A and B (Fig. 28) be two pieces of timber*
pressed in the direction of their length towards b — A by a
force equal to 6 tons weight, and B 9 tons. To find the
direction and amount of pressure they would unitedly exert,
draw the lines b e and b f in a line with the axes of the
timbers, and make b e equal to the pressure exerted by B,
viz., 9 ; also make b f equal to the pressure on A, viz., 6, and
complete the parallelogram of forces ebfd; then bd, the
diagonal of the parallelogram, will be the direction, and its
length, 9-25, will be the amount, of the united pressures of
A and of B. The line b d is termed the resultant of the two
forces b f and be. If A and B are to be supported by one
post, Ct the best position for that post will be in the direc-
tion of the diagonal bd\ and it will require to be sufficiently
strong to support the united pressures of A and of B, which
are equal to 9-25 or 9^ tons.
FIG. 29.
81.— Another Example.— Let Fig. 29 represent a piece of
Laming commonly called a crane, which is used for hoist-
ing heavy weights by means of the rope, B bf, which passes
over a pulley at b. This, though similar to Figs. 26 and 27, is,
however, still materially different. In those figures, the
strain is in one direction only, viz., from b to d\ but in this
there are two strains, from A to B and from A to W. ^ The
strain in the direction A B is evidently equal to that in the
68
CONSTRUCTION.
direction A W. To ascertain the best position for the strut
A C, make b e equal to bf, and complete the parallelogram of
forces ebfd; then draw the diagonal bd, and it will be the
position required. Should the foot, C, of the strut be placed
either higher or lower, the strain on A C would be in-
creased. In constructing cranes, it is advisable, in order
that the piece B A may be under a gentle pressure, to place
the foot of the. strut a trifle lower than where the diagonal
bd would indicate, but never higher.
W
FIG. 30.
82.— Tie§ and Strut§.— Timbers in a state of tension are
called ties, while such as are in a state of compression are
termed struts. This subject can be illustrated in the follow-
ing manner :
Let A and B (Fig. 30) represent beams of timber sup-
porting the weights W, W, and W ' ; A having but one sup-
port, which is in the middle of its length, and B two, one at
each end. To show the nature of the strains, let each beam
be sawed in the middle from a to b. The effects are obvious:
the cut in the beam A will open, whereas that in B will
close. If the weights are heavy enough, the beam A will
break at b ; while the cut in B will be closed perfectly tight
at a, and the beam be very little injured by it. But if, on
the other hand, the cuts be made in the bottom edge of the
timbers, from c to b, B will be seriously injured, while A
will scarcely be affected. . By this it appears evident that,
in a piece of timber subject to a pressure across the direction
of its length, the fibres are exposed to contrary strains. If
the timber is supported at both ends, as at B, those from the
top edge down to the middle are compressed in the direction
TIES AND STRUTS. 69
of their length, while those from the middle to the bottom
edge are in a state of tension ; but if the beam is supported
as at A, the contrary effect is produced ; while the fibres at
the middle of either beam are not at all strained. The strains
in a framed truss are of the same nature as those in a single
beam. The truss for a roof, being supported at each end,
has its tie-beam in a state of tension, while its rafters are
compressed in the direction of their length. By this, it
appears highly important that pieces in a state of tension
should be distinguished from such as are compressed, in
order that the former may be preserved continuous. A strut
may be constructed of two or more pieces ; yet, where there
are many joints, it will not resist compression so well.
83.— To Distinguish Ties from Struts.— This may be done
by the following rule. In Fig. 22-B, the timbers C and D are
the sustaining forces, and the weight £Fis the straining force ;
and if the support be removed, the straining force would
move from the point of support b towards d. Let it be
required to ascertain whether the sustaining forces arc
stretched or pressed by the straining force. Rule : Upon the
direction of the straining force b d, as a diagonal, construct
a parallelogram ebfd whose sides shall be parallel with the
direction of the sustaining forces C and D\ through the
point b draw a line parallel to the diagonal ef\ this may
then be called the dividing line between ties and struts.
Because all those supports which are on that side of the
dividing line which the straining force would occupy if
unresisted are compressed, while those on the other side of
the dividing line are stretched.
In Fig. 22-B, the supports are both compressed, being on
that side of the dividing line which the straining force would
occupy if unresisted. In Figs. 26 and 27, in which A B and
A C are the sustaining forces, A C is compressed, whereas
A B is in a state of tension ; A C being on that side of the
line h i which the straining force would occupy if unresisted,
and A B on the opposite side. The place of the latter might
be supplied by a chain or rope. In Fig. 25, the foot of the
rafter at A is sustained by two forces, the wall and the tie-
70 CONSTRUCTION.
beam, one perpendicular and the other horizontal: the
direction of the straining force is indicated by the line b a.
The dividing line h i, ascertained by the rule, shows that the
wall is pressed and the tie-beam stretched.
FIG. 31.
84.— Another Example.— Let E A B F (Fig. 31) represent
a gate, supported by hinges at A and E. In this case, the
straining force is the weight of the materials, and the direc-
tion of course vertical. Ascertain the dividing line at the
several points, G, B, /, J, H, and F. It will then appear that
the force at G is sustained by A G and G E, and the dividing
line shows that th'e former is stretched and the latter com-
pressed. The force at H is supported by A H and HE — the
former stretched and the latter compressed. The force at B
is opposed by H B and A B, one pressed, the other stretched.
The force at F is sustained by G F and FE, G F being
stretched and F E pressed. By this it appears that A B is in
a state of tension, and E F of compression ; also, that A H
and G F are stretched, while B H and G E are compressed :
which shows the necessity of having A H and G F each in
one whole length, while B H and G E may be, as they are
shown, each in two pieces. The force at J is sustained by
GJ and J H, the former stretched and the latter compressed.
The piece C D is neither stretched nor pressed, and could
be dispensed with if the joinings at J and / could be made
TO FIND THE CENTRE OF GRAVITY. 71
as effectually without it. In case A B should fail, then C D
would be in a state of tension.
85. — Centre of Gravity. — The centre of gravity of a uni-
form prism or cylinder is in its axis, at the middle of its
length ; that of a triangle is in a line drawn from one angle
to the middle of the opposite side, and at one third of the
length of the line from that side ; that of a right-angled tri-
angle, at a point distant from the perpendicular equal to one
third of the base, and distant from the base equal to one
third of the perpendicular ; that of a pyramid or cone, in
the axis and at one quarter of the height from the base.
The centre of gravity of a trapezoid (a four-sided, figure
having only two of its sides parallel) is in a line joining the
centres of the two parallel sides, and at a distance from the
longest of the parallel sides equal to the product of the
length in the sum of twice the shorter added to the longer
of the parallel sides, divided by three times the sum of the
two parallel sides. Algebraically thus :
,_'_(2_
- 3 (« +
where d equals the distance from the longest of the parallel
sides, / the length of the line joining the two parallel sides,
and a the shorter and b the longer of the parallel sides.
Example. — A rafter 25 feet long has the larger end 14
inches wide, and the smaller end 10 inches wide: how far
from the larger end is the centre of gravity located ?
Here / = 25, a — |f, and b = |f,
- _ . 25_x_ = 25x34,
" W±P) ' 3(W + «) " 3xff 3x24 -
— ° =11-8 = 11 feet 9! inches nearly.
In irregular bodies with plain sides, the centre of gravity
may be found by balancing them upon the edge of a prism
— upon the edge of a table — in two positions, making a line
each time upon the body in a line with the edge of the prism,
and the intersection of those lines will indicate the point re-
72 CONSTRUCTION.
quired. Or suspend the article by a cord or thread attached
to one corner or edge ; also from the same point of suspen-
sion hang a plumb-line, and mark its position on the face of
the article; again, suspend the article from another corner
or side (nearly at right angles to its former position), and
mark the position of the plumb-line upon its face ; then the
intersection of the two lines will be the centre of gravity.
FIG. 32.
86. — Effect of the Weight of Inclined Beam§. — An in-
clined post or strut supporting some heavy pressure applied
at its upper end, as at Fig. 25, exerts a pressure at its foot in
the direction of its length, or nearly so. But when such a
beam is loaded uniformly over its whole length, as the rafter
of a roof, the pressure at its foot varies considerably from
the direction of its length. For example, let A B (Fig. 32)
be a beam leaning against the wall B c, and supported at its
foot by the abutment A, in the beam A c, and let o be the
centre of gravity of the beam. Through o draw the verti-
cal line b dy and from B draw the horizontal line B b, cutting
b dm b\ join b and A, and b A will be the direction of the
thrust. To prevent the beam from loosing its footing, the
joint at A should be made at right angles to b A. The
amount of pressure will be found thus : Let b d (by any scale
of equal parts) equal the number of tons upon the beam
A B\ draw d e parallel to B b ; then /; c (by the same scale)
equals the pressure in the direction b A ; and e d the pres-
sure against the wall at B— and also the horizontal thrust at
A^ as these are always equal in a construction of this kind.
The horizontal thrust of an inclined beam (Fig. 32)— the
effect of its own weight— may be calculated thus :
Rule.— Multiply the weight of the beam in pounds by
THRUST OF INCLINED BEAMS. 73
its base, A C, in feet, and by the distance in feet of its centre
of gravity, o (see Art. 85), from the lower end, at A, and
divide this product by the product of the length, A B, into
the height, B C, and the quotient will be the horizontal
thrust in pounds. This may be stated thus : H = — ,
where d equals the distance of the centre of gravity, 0, from
the lower end ; b equals the base, A C ; iv equals the weight
of the beam ; h equals the height, D C ; /equals the length
of the beam ; and H equals the horizontal thrust.
Example. — A beam 20 feet long weighs 300 pounds; its
centre of gravity is at 9 feet from its lower end ; it is so in-
clined that its base is 16 feet and its height 12 feet : what is
the horizontal thrust ?
TT d b w , o x 16 x 300 Q x 4 x 25
Here — — — becomes — — — - — — 0x4x5
hi 12x20 5
= 180 =H =. the horizontal thrust.
This rule is for cases where the centre of gravity does
not occur at the middle of the length of the beam, although
it is applicable when it does occur at the middle ; yet a
shorter rule will suffice in this case, and it is thus:
Rule. — Multiply the weight of the rafter in pounds by
the base, A C (Fig. 32), in feet, and divide the product by
twice the height, B C, in feet, and the quotient will be the
horizontal thrust, when the centre of gravity occurs at the
middle of the beam.
If the inclined beam is loaded with an equally distributed
load, add this load to the weight of the beam, and use this
total weight in the rule instead of the weight of the beam.
And generally, if the centre of gravity of the combined
weights of the beam and load does not occur at the centre
of the length of the beam, then the former rule is to be used.
In Fig. 33, two equal beams are supported at their feet by
the abutments in the tie-beam. This case is similar to the
last ; for it is obvious that each beam is in precisely the
position of the beam in Fig. 32. The horizontal pressures at
B, being equal and opposite, balance one another ; and their
horizontal thrusts at the tie-beam are also equal. (See Art.
CONSTRUCTION.
78— Fig. 25.) When the height of a roof (Fig. 33) is one
fourth of the span, or of a shed (Fig. 32) is one halt the
span, the horizontal thrust of a rafter, whose centre of grav-
FIG. 33.
ity is at the middle of its length, is exactly equal to the
weight distributed uniformly over its surface.
In shed or lean-to roofs, as Fig. 32, the horizontal pressure
will be entirely removed if the bearings of the rafters, as A
and B (Fig. 34), are made horizontal — provided, however,
FIG. 34.
that the rafters and other framing do not bend between the
points of support. If a beam or rafter have a natural curve,
the convex or rounding edge should be laid uppermost.
87. — Effect of Load 011 Beam. — The strain in a uniformly
loaded beam, supported at each end, is greatest at the
middle of its length. Hence mortices, large knots, and other
defects should be kept as far as possible from that point ;
and in resting a load upon a beam, as a partition upon a
floor-beam, the weight should be so adjusted, if possible,
that it will bear at or near the ends.
Twice the weight that will break a beam, acting at the
centre of its length, is required to break it when equally
VARYING PRESSURE ON BEARINGS. 75
distributed over its length ; and precisely the same deflec-
tion or .rag- 'will be produced on a beam by a load equally
distributed that five eighths ot the load will produce if act-
ing at the centre of its length.
88. — Effect 011 Bern-ings. — When a uniformly loaded
beam is supported at each end on level bearings (the beam
itself being either horizontal or inclined), the amount of
pressure caused by the load on each point of support is
equal to one half the load ; and this is also the ase when
the load is concentrated at the middle of the beam, or has
its centre of gravity at the middle of the beam ; but when
the load is unequally distributed, or concentrated so that its
centre of gravity occurs at some other point than the middle
of the beam, then the amount of pressure caused by the
load on one of the points of support is unequal to that on
the other. The precise amount on each may be ascertained
by the following rule.
Rule.— Multiply the weight W (Fig. 35) by its distance,
C B, from its nearest point of support, By and divide the pro-,
duct by the length, A B, of the beam, and the quotient will
FIG 35.
be the amount of pressure on the remote point of support, A.
Again, deduct this amount from the weight W, and the re-
mainder will be the amount of pressure on the near point of
support, B ; or, multiply the weight W by its distance, A C,
from the remote point of support, A, and divide the pro-
duct by the length, A B, and the quotient will be the amount
of pressure on the near point of support, B.
When / equals the length between the bearings A and B,
n — AC, m — C B, and W — the load ; then
7g CONSTRUCTION.
v/ I
JL_/- — A = the amount of pressure at A, and
. ? = ,5 = the amount of pressure at ^?.
Example.— A beam 20 feet long between the bearings
has a load of 100 pounds concentrated at 3 feet from one of
the bearings : what is the portion of this weight sustained by
each bearing ?
Here W— 100 ; «, 17 ; m, 3 ; and /, 20.
W m
Hence A=-j-
Load on A = 15 pounds.
Load on ^ = 85 pounds,
Total weight = 100 pounds.
RESISTANCE OF MATERIALS.
89. — Weight— Strength. — Preliminary to designing a roof-
truss or other piece of framing, a knowledge of two subjects
is essential : one is, the effect of gravity acting upon the
various parts of the intended structure ; the other, the power
of resistance possessed by the materials of which the framing
is to be constructed. The former subject having been
treated of in the preceding pages, it remains now to call at-
tention to the latter.
90. — Quality of Materials.— Materials used in construc-
tion are constituted in their structure either of fibres
(threads) or of grains, and are termed, the former fibrous,
the latter granular. All woods and wrought metals are
fibrous, while cast iron, stone, glass, etc., are granular. The
strength of a granular material lies in the power of attrac-
tion acting among the grains of matter of which the mate-
rial is composed, by which it resists any attempt to separate
its grains or particles of matter. A fibre of wood or ot
UNIVERSITY
THE THREE KINDS OF ST
wrought metal has a strength by which it resists being com-
pressed or shortened, and finally crushed ; also a strength
by which it resists being extended or made longer, and
finally sundered. There is another kind of strength in a
fibrous material : it is the adhesion of one fibre to another
along their sides, or the lateral adhesion of the fibres.
91. — Manner of Rc§isting. — In the strain applied to a
post supporting a weight imposed upon it (Fig. 36), we
have an instance of an essay to shorten the fibres of which
the timber is composed. The strength of the timber in
this case is termed the resistance to compression. In the strain
on a piece of timber like a king-post or suspending piece
(A, Fig. 37), we have an instance of an essay to extend or
lengthen the fibres of the material. The strength here ex-
hibited is termed the resistance to tension. When a piece of
timber is strained like a floor-beam or any horizontal piece
FIG 37.
FIG 38.
carrying a load (Fig. 38). we have an instance in which the
two strains of compression and tension are both brought
into action ; the fibres of the upper portion of the beam be-
ing compressed, and those of the under part being stretched.
78 CONSTRUCTION.
This kind of strength of timber is termed resistance to cross-
strains. In each of these three kinds of strain to which tim-
ber is subjected, the power of resistance is in a measure due
to the lateral adhesion of the fibres, not so much perhaps in
the simple tensile strain, yet to a considerable degree in the
compressive and cross strains. But the power of timber,
by which it resists -a pressure acting compressively in the
direction of the length of the fibres, tending to separate the
timber by splitting off a part, as in the case of the end of a
tie-beam, against which the foot of the rafter presses, is
wholly due to the lateral adhesion of the fibres.
92. — Strength and Stiffne§§. — The strengtJi of materials
is their power to resist fracture, while the stiffness of mate-
rials is their capability to resist deflection or sagging. A
knowledge of their strengtJi is useful, in order to determine
their limits of size to sustain given weights safely ; but a
knowledge of their stiffness is more important, as in almost
all constructions it is desirable not only that the load be safely
sustained, but that no appearance of weakness be manifested
by any sensible deflection or sagging.
93. — Experiments : Constants — In the investigation of
the laws applicable to the resistance of materials, it is found
that the dimensions — length, breadth, and thickness — bear
certain relations to the weight or pressure to which the
piece is subjected. These relations are general ; they exist
quite independently of the peculiarities of any specific piece
of material. These proportions between the dimensions
and the load are found to exist alike in wood, metal, stone,
and glass, or other material. One law applies alike to all
materials ; but the capability of materials to resist differs in
accordance with the compactness and cohesion of particles,
and the tenacity and adhesion of fibres, those qualities upon
which depends the superiority of one kind of material over
another. The capability of each particular kind of material
is ascertained by experiments, made upon several specimens,
and an average of the results thus obtained is taken as an
index of the capability of that material, and is introduced
in the rules as a constant number, each specific kind of ma-
VALUES OF WOODS FOR COMPRESSION.
79
terial having- its own special constant, obtained by ex-
perimenting- on specimens of that peculiar material. The
results of experiments made to test the resistance of various
materials useful in construction — their capability to resist
the three strains before named — will now be introduced.
94. — Resistance to Compression. — The following table
exhibits the results of experiments made to test the resist-
ance to compression of such woods as are in common use in
this country for the purposes of construction.
TABLE I. — RESISTANCE TO COMPRESSION.
MATERIAL.
>,
">
Z
O
u
te
1
To crush fibres ~
longitudinally. •
To separate fibres
by sliding. fS
To crush fibres trans-
versely 3^ inch deep.
Value of .Pin Rules.
Sensible Impres- ^
sion.
Georgia Pine
0-611
Pounds
per inch.
95OO
Pounds
per inch.
840
Pounds
per inch.
225O
QOO
0-762
II7OO
1160
28OO
1 1 2O
White Oak
O- 774
8000
I25O
2650
IO6O
Spruce
0-360
7850
540
650
26O
White Pine
O-^88
6650
480
800
32O
Hemlock
O-423
57OO
370
800
32O
White Wood
o- 307
3400
800
320
Chestnut
O-4QI
6700
1250
5OO
Ash
o- 517
5850
3'OO
1240
Maple
O-574
8450
2700
I080
Hickory
O-877
13750
4100
1640
Cherry
O*4Q4
QO5O
2500
IOOO
Black Walnut
O-42I
7800
2IOO
840
Mahogany (St. Domingo)
(Bay Wood)
0-837
O-43Q
Il6oO
4900
•
5700
I7OO
2280
680
Live Oak
o • 016
IIIOO
6800
2720
Lignum Vita? . .
\J yj.w
1-282
I2IOO
7700
3080
The resistance of timber of the same name varies much ;
depending as it obviously must on the soil in which it grew
on its age before and after cutting, on the time of year
when cut, and on the manner in which it has been kept since
it was cut. And of wood from the same tree much depends
upon its location, whether at the butt or towards the limbs,
and whether at the heart or of the sap, or at a point mid-
way from the centre to the circumference of the tree. The
8O CONSTRUCTION.
pieces submitted to experiment were of ordinary good
quality, such as would be deemed proper to be used in
framing. The prisms crushed were generally small, about
2 inches long, and from I inch to i-J inches square ; some
were wider one way than the other, but all containing in
area of cross section from I to 2 inches. The weight given
in the table is the average weight per superficial inch.
Of the first six woods named, there were nine specimens
of each tested ; of the others, generally three specimens.
The results for the first six woods named are taken
from the author's work on Transverse Strains, published by
John Wiley & Sons, New York. The results for these six
woods, as well as those for all the others named in the table,
were obtained by experiments carefully made by the author.
The first six woods named were tested in 1874 and 1876, and
upon a testing machine, in which the power is transmitted
to the pieces tested, by levers acting upon knife-edges.
For a description of this machine, see Transverse Strains,
Art. 704. The woods named in the table, other than the
first six, were tested some twenty years since, and upon a
hydraulic press, which, owing to friction, gave results too
low.
The results, as thus ascertained, were given to the public
in the 7th edition of this work, in 1857. In the present edi-
tion, the figures in Table I., for these woods, are those
which have resulted by adding to the results given by the
hydraulic press a certain quantity thought to be requisite
to compensate for the loss by friction. Thus corrected, the
figures in the table may be taken as sufficiently near approx-
imations for use in the rules, — although not so trustworthy
as the results given for the first six woods named, as these
were obtained upon a superior testing machine, as above
stated.
In the preceding table, the second column contains the
specific gravity of the several kinds of wood, showing their
comparative density. The weight in pounds of a cubic foot
of any kind of wood or other material is equal to its
specific gravity multiplied by 62-5, this number being the
weight in pounds of a cubic foot of water. The third column
EXPLANATION OF TABLE I. 8 1
contains the weight in pounds required to crush a prism
having a base of one inch square ; the pressure applied to
the fibres longitudinally. In practice, it is usual never to
load material exposed to compression with more than one
fourth of the crushing weight, and generally with from one
sixth to one tenth only. The fourth column contains the
weight in pounds wrhich, applied in line with the length of the
fibres, is required to force off a part of the piece, causing the
fibres to separate by sliding, the surface separated being one
inch square. The fifth column contains the weight in pounds
required to crush the piece when the pressure is applied to the
fibres transversely, the piece being one inch thick, and the
surface crushed being one inch square, and depressed one
twentieth of an inch deep. The sixth column contains the
value of P in the rules ; P being the weight in pounds, ap-
plied to the fibres transversely, which is required to make a
sensible impression one inch square on the side of the piece,
this being the greatest weight that would be proper for a
post to be loaded with per inch surface of bearing, resting
on the side of the kind of wood set opposite in the table. A
greater weight would, in proportion to the excess, crush the
side of the wood under the post, and proportionably derange
the framing, if not cause a total failure. It will be observed
that the measure of this resistance is useful in limiting the
load on a post according to the kind of material contained,
not in the post, but in the timber upon which the post presses.
95. — He§i§tance to Tension. — The resistance of materials
to the force of stretching, as exemplified in the case of a
rope from which a weight is suspended, is termed the resist-
ance to tension. In fibrous materials, this force will be differ-
ent in the same specimen, in accordance \vith the direction
• in which the force acts, whether in the direction of the length
of the fibres or at right angles to the direction of their length.
It has been found that, in hard woods, the resistance in the
former direction is about eight to ten times what it is in the
latter; and in soft woods, straight, grained, such as white
pine, the resistance is from sixteen to twenty times. A
knowledge of the resistance in the direction of the 1
the most useful in practice.
82
CONSTRUCTION.
In the following table are recorded the results of ex-
periments made to test this resistance in some of the woods
in common use, and also in iron, cast and wrought. Each
specimen of the woods was turned cylindrical, and about 2
inches diameter, and then the middle part reduced to about
f of an inch diameter, at the middle of the reduced part,
and thence gradually increased toward each end, where it
was considerably larger at its junction with the enlarged
end. The results, in the case of the iron and of the first six
woods named, are taken from the author's work, Trans-
verse Strains, Table XX. Experiments were made upon
the other three woods named by a hydraulic press, some
twenty years since, and the results were first published in
the 7th edition of this work, in 1857. These results, owing
to friction, were too low. Adding to them what is supposed
to be the loss by friction of the machine, the results thus
corrected are what are given for these three woods in the
following table, and may be taken as fair approximations,
but are not so trustworthy as the figures given for the other
six woods and for the metals.
TABLE II. — RESISTANCE TO TENSION.
MATERIAL.
Specific
Gravity.
T.
Pounds re-
quired to rup-
ture one inch
square.
Georgia Pine
O-6^
16000
Locust
O'7Q4
24800
White Oak
O • 77J.
IQCQO
O 4^2
TOCOO
White Pine
Q. 4^8
I2OCO
Hemlock f
o • 402
87OO
Hickory
O- 7^1
26OCO
Maple
O- 6o4
2OOOO
Ash
O-6o8
I5OOO
Cast Iron, American ) from
6 • Q44
English f to
7 • c84
1 7OOO
Wrought Iron, American ) from
7 • 6OO
English y to
7 • 7Q2
5OOOO
The figures in the table denote the ultimate capability of
a bar one inch square, or the weight in pounds required to
VALUES OF MATERIALS FOR CROSS-STRAINS.
produce rupture. Just what portion of this should be taken
as the safe capability will depend upon the nature of the
strain to which the material is to be exposed. In practice it
is found that, through defects in workmanship, the attach-
ments may be so made as to cause the strain to act along one
side of the piece, instead of through its axis; and that in this
case fracture will be produced with one third of the strain
that can be sustained through the axis. Due to this and
other contingencies, it is usual to subject materials exposed
to tensile strain with only from one sixth to one tenth of
the breaking weight.
96. — Resistance to Transverse Strains. — In the follow-
ing table are recorded the results of experiments made to
test the capability of the various materials named to resist
the effects of transverse strain. The figures are taken from
the author's work, Transverse Strains, before referred to.
TABLE III. — TRANSVERSE STRAINS.
MATERIAL.
*
Resistance
to
Rupture.
F.
Resistance
to
Flexure.
e.
Extension
of
Fibres.
a,
Margin
for
Safety.
Georgia. Pine
850
5QOO
•00109
I-84
I2OO
5050
•OOI5
2-20
White Oak
6"iO
3IOO
•00086
3'39
55°
35OO
• 00098
2-23
White Pine
500
2900
•0014
I-7I
Hemlock
450
2800
•00095
2-35
White Wood
600
3450
•00096
2-52
480
2550
•00103
2-54
A«*h
QOO
4OOO
•OOIII
2 82
Maple
IIOO
5I50
• 0014
2-12
1050
3850
•0013
2-QI
Cherry
650
2850
•001563
2-03
Black Walnut
750
3900
•00104
2-57
Mahogany (St Domingo) ....
650
3600
00116
2-16
(Bay Wood)
850
475°
•00109
2-28
Cast Iron, American . .
" English
2500
2IOO
50000
40000
26OO
62000
" English
iqoo
60000
Steel in Bars •
6000
70000
1
200
59
33
' ' pressed
37
Marble East Chester
147
84 CONSTRUCTION.
The figures in the second column, headed B, denote the
weight in pounds required to break a unit of the material
named when suspended from the middle, the piece being
supported at each end. The unit of material is a bar one inch
square and one foot long between the bearings. The third
column, headed F, contains the values of the several mate-
rials named as to their resistance to flexure, as explained in
Arts. 302-305, Transverse Strains. These values of F, as
constants, are used in the rules. The fourth column, headed
^, contains the values of the several materials named, denot-
ing the elasticity of the fibres, as explained in Art. 312,
Transverse Strains. These values of e, as constants, are
used in the rules.
The fifth column, headed ay contains for the several ma-
terials named the ratio of the resistance to flexure as com-
pared with that to rupture, and which, as constants used
in the rules, indicate the margin of safety to be given for
each kind of material. The figures given in each case show
the smallest possible value that may be safely given to a, the
factor of safety, x In practice it is generally taken higher
than the amount given in the table. For example, in the
table the value of B, the constant for rupture by transverse
strain for spruce, is 550.
Now, if the dimensions of a spruce beam to carry a given
weight be computed by the rules, using the constant B, at
550, the beam will be of such a size that the given weight
will just break it.
But if, in the computation, instead of taking the full
value of B, only a part of it be taken, then the beam will not
break immediately; and if the part taken be so small that
its effect upon the fibres shall not be sufficient to strain them
beyond their limit of elasticity, the beam will be capable of
sustaining the weight for an indefinite period ; in this case
the beam will be loaded by what is termed the safe weight.
Or, since the value of a for spruce is 2-23 in the table, if, in-
stead of taking B at 550, its full value, only the quotient
arising from a division of B by a be taken — or 550 divided
by 2-23, which equals 246-6 — then the beam will be of suffi-
cient size to carry the load safely. Therefore, while the con-
THE VARIOUS CLASSES OF PRESSURES. 85
slant B is to be used for a breaking weight, for a safe load
n
the quotient of - is to be used. But, again, if a be taken at
its value as given in the table, the computed beam will be
loaded up to its limit of safety. So loaded that, if the load
be increased only in a small degree, the limit of safety will
be passed, and the beam liable, in time, to fail by rupture.
Therefore, as the values of a, in the table, are the smallest
possible, it is prudent in practice always to take a larger
than the table value. For example, the table value of
a for spruce is 2-23, but in practice let it be taken at 3
or 4.
97. — Resistance to Compression. — The resistance of ma-
terials to the force of compression may be considered in
several ways. Posts having their heights less than ten
times their least sides will crush before bending ; these
belong to one class : another class is that which com-
prises all posts the height of which is equal to ten times
their least sides, or more than ten times ; these will bend
before crushing. Then there remains to be considered
the manner in which the pressure is applied : whether in
line with the fibres, or transversely to them ; and, again,
whether the pressure tends to crush the fibres, or simply
to force off a part of the piece by splitting. The various
pressures may be comprised in the four classes following,
namely :
ist. — When the pressure is applied to the fibres trans-
versely.
2d.— When the pressure is applied to the fibres longi-
tudinally, and so as to split off the part pressed against,
causing the fibres to separate by sliding.
3d. — When the pressure is applied to the fibres longi-
tudinally, and on short pieces.
4th.— When the pressure is applied to the fibres longi-
tudinally, and on long pieces.
These four classes will now be considered in their reg-
ular order.
86 CONSTRUCTION.
98. — Compression Transversely to the Fibres. — In this
first class of compression, experiment has shown that the
resistance is in proportion to the number of fibres pressed,
that is, in proportion to the area. For example, if 5000
pounds is required to crush a prism with a base i inch
square, it will require 20,000 pounds to crush a prism having
a base of 2 by 2 inches, equal to 4 inches area ; because 4
times 5000 equals 20000.
Therefore, if any given surface pressed be multiplied by
the pressure per inch which the kind of material pressed
may be safely trusted with, the product will be the total
pressure which may with safety be put upon the given sur-
face. Now, the capability for this kind of resistance is given
in column P, in Table I., for each kind of material named in
the table. Therefore, to find the limit of weight, proceed
as follows:
99. — The Limit of Weight. — To ascertain what weight
a post may be loaded with, so as not to crush the surface
against which it presses, we have —
Rule 1.— Multiply the area of the post in inches by the
value of P, Table I., and the product will be the weight re-
quired in pounds ; or —
W=AP. (i.)
Example. — A post, 8 by 10 inches, stands upon a white-
pine girder; the area equals 8 x 10 = 80 inches. This being
multiplied by 320, the value of P, Table I., set opposite white
pine, the product, 25600, is the required weight in pounds.
100. — Area of Post.— To ascertain what area a post must
have in order to prevent the post, loaded with a given
weight, from crushing the surface against which it presses,
we have —
Rule II. — Divide the given weight in pounds by the value
of P, Table I., and the quotient will be the area required in
inches ; or--
RESISTANCE TO RUPTURE BY SLIDING. 87
Example. — A post standing on a Georgia-pine girder is
loaded with 100,000 pounds: what must be its area? The
weight, 100000, divided by 900, the value of /> Table I., set
opposite Georgia pine, the quotient, in- 11, is the required
area in inches. The post may be 10 by ii|, or 10 by 11
inches; or if square each side will be 10-54 inches, or IIT'¥
inches diameter if round.
101.— Rupture by Sliding. — In this the second class of
rupture by compression, it has been ascertained by ex-
periment that the resistance is in proportion to the area of
the surface separated without regard to the form of the sur-
face. Now, in Table I., column //, we have the ultimate
resistance to this strain of the several materials named.
But to obtain the safe load per inch, the ultimate resist-
ance of the table is to be divided by a factor of safety, of
such value as circumstances may seem to require. Gener-
ally this factor may be taken at 3. Then to obtain the safe
load for any given case, we have but to multiply the given
surface by the ultimate resistance, and divide by the factor
of safety ; therefore, proceed as follows :
102.— Tl»e Limit of Weight.— To ascertain what weight
may be sustained safely by the resistance of a given area of
surface, when the weight tends to split off the part pressed
against by causing, in case of fracture, one surface to slide
on the other, we have —
Rule III.— Multiply the area of the surface by the value
of Hi in Table I. divide by the factor of safety, and the
quotient will be the weight required in pounds ; or—
W = -*-Z (3-)
Example. — The foot of a rafter is framed into the end of
its tie-beam, so that the uncut substance of the tie-beam is
1 5 inches long from the end of the tie-beam to the joint of
the rafter; the tie-beam is of white pine, and is 6 inches
thick: what amount of horizontal thrust will this end of
the tie-beam sustain, without danger of having the end oi
88 CONSTRUCTION.
the tie-beam split off? Here the area of surface that sus-
tains the pressure is 6 by 15 inches, equal to 90 inches.
This multiplied by 480, the value of H, set opposite to
white pine, Table I., and divided by 3, as a factor of safety,
gives a quotient of 14400, and this is the required weight in
pounds.
103. — Area of Surface. — To ascertain the area of surface
that is required to sustain a given weight safely, when the
weight tends to split off the part pressed against, by causing,
in case of fracture, one 'surface to slide on the other, we
have—
Rule IV. — Divide the given weight in pounds by the
value of H, Table I. ; multiply the quotient by the factor of
safety, and the product will be the required area in inches ;
or —
(40
Example. — The load on a rafter causes a horizontal thrust
at its foot of 40,000 pounds, tending to split off the end of
the tie-beam : what must be the length of the tie-beam be-
yond the line where the foot of the rafter is framed into it,
the tie-beam being of Georgia pine, and 9 inches thick ?
The weight, or horizontal thrust, 40000, divided by 840, the
value of //, Table I., set opposite Georgia pine, gives a quo-
tient of 47-619, and this multiplied by 3, as a factor of safety,
gives a product of 142-857. This, the area of surface in
inches, divided by 9, the breadth of the surface strained
(equal to the thickness of the tie-beam), the quotient, 15.87,
is the length in inches from the end of the tie-beam to the
rafter joint, say 16 inches.
I04.-Tcnon§ and Splices.- A knowledge of this kind of
resistance of materials is useful, also, in ascertaining the
length of framed tenons, so as to prevent the pin, or key,
with which they are fastened from tearing out ; and, also, in
cases where tie-beams, or other timber under a tensile strain,
CRUSHING STRENGTH OF STOUT POSTS. 89
are spliced, this rule gives the length of the joggle at each
end of the splice.
105. — Stout Post§.— These comprise the third class of ob-
jects subject to compression (Art. 97), and include all posts
which are less than ten diameters high. The resistance to
compression, in this class, is ascertained to be directly in pro-
portion to the area of cross-section of the post.
Now in Table I., column Cy the ultimate resistance to
crushing is given for the several kinds of materials named ;
from which the safe resistance per inch may be obtained by
dividing it by a proper factor of safety. Having the safe
resistance per inch, the resistance of any given post may be
determined by multiplying it by the area of the cross-section
of the post. Therefore, proceed as follows :
106.— TBie Limit of Weight.— To find the weight that can
be safely sustained by a post, when the height of the post is
less than ten times the diameter if round, or ten times the
thickness if rectangular, and the direction of the pressure
coinciding with the axis, we have—
RHle V. — Multiply the area of the cross-section of the
post in inches by the value of C, in Table I. ; divide the pro-
duct by the factor of safety, and the quotient will be the re-
quired weight in pounds ; or —
W= ^-£. (5.)
Example.— A Georgia-pine post is 6 feet high, and in
cross-section 8 x 12 inches: what weight will it safely sus-
tain? The height of this post, 12 x 6 = 72 inches, which is
less than lox 8 (the size of the narrowest side) — 80 inches ;
it therefore belongs to the class coming under this rule. The
area = 8 x 12 = 96 inches ; this multiplied by 9500, the value
of C, in the table, set opposite Georgia pine, and divided by
6, as a factor of safety, the quotient, 152000, is the weight in
pounds required. It will be observed that the weight would
be the same for a Georgia-pine post of any height less than
90 CONSTRUCTION.
10 times 8 inches = So inches = 6 feet 8 inches, provided its
breadth and thickness remain the same, 12 and 8 inches.
107. — Area of Post. — To find the area of the cross-sec-
tion of a post to sustain a given weight safely, the height of
the post being less than ten times the diameter if round, or
ten times the least side if rectangular, the pressure coinciding
with the axis, we have—
Rule VI. — Divide the given weight in pounds by the
value of C, in Table I. -, multiply the quotient by the factor
of safety, and the product will be the required area in
inches ; or—
(6.)
Example. — A weight of 40,000 pounds is to be sustained
by a white-pine post 4 feet high : what must be its area of
section to sustain the weight safely ? Here, 40000 divided
by 6650, the value of C, in Table I., set opposite white pine,
and the quotient multiplied by 6, as a factor of safety, the pro-
duct is 36 ; this, therefore, is the required area, and such a
post may be 6 x 6 inches. To find the least side, so that it
shall not be less than one tenth of the height, divide the
height, reduced to inches, by 10, and make the least side to
exceed this quotient. The area divided by the least side so
determined will give the wide side. If, however, by this
process, the first side found should prove to be the greatest,
then the size of the post is to be found by Rule IX., X., or
XI.
108. — Area of Round Po§t§.— In case the post is to be
round, its diameter may be found by reference to the Table
of Circles in the Appendix, in the column of diameters, op-
posite to the area of the post to be found in the column of
areas, or opposite to the next nearest area. For example,
suppose the required area, as just found by the example
under Rule VI., is 36 : by reference to the column of areas,
35.78 is the nearest to 36, and the diameter set opposite is
CRUSHING STRENGTH OF SLENDER POSTS. 9!
6.75, which is a trifle too small. The post may therefore be,
say, 6| inches diameter.
109. — Slender Posts. — When the height of a post is less
than ten times its diameter, the resistance of the post to
crushing is approximately in proportion to its area of cross-
section. But when the height is equal to or more than ten
diameters, the resistance per square inch is diminished. The
resistance diminishes as the height is increased, the diameter
remaining the same (Transverse Strains, Art. 643). The
strength of a slender post consists in a combination of the
resistances of the material to bending and to crushing, and
is represented in the following rule :
110. — The Limit of Weight. — To ascertain the weight
that can be sustained safely by a post the height of which
is at least ten times its least side if rectangular, or ten times
its diameter if round, the direction of the pressure coincid-
ing with the axis, we have —
Rule VII. — Divide the height of the post in inches by the
diameter, or least side, in inches ; multiply the quotient by
itself, or take its square ; multiply the square by the value
of e, in Table III., set opposite the kind of material of which
the post is made ; to the product add the half of itself ; to
the sum add unity (or one) ; multiply this sum by the factor
of safety, and reserve the product for use, as below. Now
multiply the area of cross-section of the post in inches by
the value of C, in Table I., set opposite the material of the
post, and divide the product by the above reserved product;
the quotient will be the required weight in pounds ; or—
(7.)
Example : A Round Post.— What weight may be safely
placed upon a post of Georgia pine 10 inches diameter and
10 feet high, the pressure coinciding with the axis of the
post? The height of the post, dox 12 =) 120 inches, divided
by 10, its diameter, gives a quotient of 12; this multiplied
92 CONSTRUCTION.
by itself gives 144, its square; and this by -00109, the value
of e for Georgia pine, in Table III., gives • 15696 ; to which
adding its half, the sum is 0-23544; to which adding unity,
the sum is 1-23544 ; and this multiplied by 7, as a factor of
safety, the product is 8 -648, the reserved divisor. Now the
area of the post is (see Table of Areas of Circles, in the Ap-
pendix, opposite its diameter, 10) 78-54; this multiplied by
9500, the value of C for Georgia pine, in Table I., gives a
product of 746130; which divided by 8-648, the above re-
served divisor, gives a quotient of 86278, the required weight
in pounds.
Anotlier Example : A Rectangular Post. — What weight may
be safely placed upon a white-pine post lox 12 inches, and
15 feet high, the pressure coinciding with the axis of the
post? Proceeding according to the rule, we find the height
of the post to be 180 inches, which divided by 10, the least
side of the post, gives 18 ; this multiplied by itself gives 324*
its square ; which multiplied by -0014, the value of e for
white pine, in Table III., gives -4536; to which adding its
half, the sum is -6804; to which adding unity, the sum is
i -6804 ; and this multiplied by 8, as a factor of safety, the pro-
duct is 13-4432, the reserved divisor. Now the area of the
post, ( 10 x 12 =) 120 inches, multiplied by 6650, the value of
C for white pine, in Table I., gives a product of 798,000, and
this divided by 13-4432, the above reserved divisor, the quo-
tient, 59360, is the required weight in pounds.
III. — Diameter of the Post: when Round.— To ascertain
the size of a round post to sustain safely a given weight,
when the height of the post is at least ten times the diameter ;
the direction of the pressure coinciding with the axis of the
post; we have —
Rule VIII. —Multiply the given weight by the factor of
safety, and divide the product by 1-5708 times the value of C
for the material of the post, found in Table I. ; reserve the
quotient, calling its value G. Now multiply 432 times the
value of c for the material of the post, found in Table III.,
by the square of the height in feet, and by the above quo-
tient G ; to the product add the square of G : extract the
SIZE OF POST FOR GIVEN WEIGHT. 93
square root of the sum, and to it add the value of G ; then
the square root of this sum will be the required diameter;
or—
r Wa
Cr nzr
,5708 L
d =4 / . / At? (T~f /8 O. f^ 4- /C * (90
Example. — What should be the diameter of a locust post
10 feet high to sustain safely 40,000 pounds, the pressure
coinciding with the axis ? Proceeding by the rule, the given
weight multiplied by 6, taken as a factor of safety, equals
240000. Dividing this by 1-5708 times 11700, the value of
C for locust, in Table I., the quotient, 13-06, is the value of
G, the square of which is 170-53. Now, the value of e for
locust, in Table III., is -0015. This multiplied by 432, by
100, the square of the height, and by the above value of G,
gives a product of 846-2 ; which added to 170-53, the above
square of G, gives the sum of 1016-73. To 31-89, the square
root of this, add the above value of G ; then 6-7, the square
root of this sum, is the required diameter of the post. The
post therefore requires to be 6-7, say 6£- inches diameter.
112.— Side of tlic Post: when Square.— To ascertain the
side of a square post to sustain safely a given weight, when
the height of the post is at least ten times the side ; the pres-
sure coinciding with the axis ; we have —
Rule IX.— Multiply the given weight by the factor of
safety, and divide the product by twice the value of C for
the material of the post, found in Table I. ; reserve the quo-
tient, calling its value G.. Now multiply 432 times the value
of e for the material of the post, found in Table III., by the
square of the height in feet, and by the above quotient G\
to the product add the square of G ; extract the square root
of the sum, and to it add the value of G ; then the square
root of this sum will be the required side of the post ; or—
. Co-)
2 C
94 CONSTRUCTION.
•S =4/ 4/432 Ge I * + G'2 + G.
Example. — What should be the side of a Georgia-pine
square post 1 5 feet high to sustain safely 50,000 pounds, the
pressure coinciding with the axis of the post? Proceeding
by the rule, 50,000 pounds multiplied by 6, as a factor of
safety, gives 300000 ; this divided by 2 x 9500 (the value of
€)•=• 19000, the quotient, 15-789, is the value of G. The
value of e for Georgia pine is -00109; tne square of the
height is 225 ; then, 432 times -00109 by 225 and by 15-789
(the above value of G) gives a product of 1672 - 86 ; the square
of £ equals 249-31 ; this added to 1672-86 gives a sum of
1922- 17, the square root of which is 43-843 ; which added* to
15-789, the value of G, gives 59-632, the square root of which
is 7-722, the required side of the post. The post, therefore,
requires to be, say, 7f inches square.
113. — Thickness of a Rectangular Post.— This may be
definitely ascertained when the proportion which the thick-
ness shall bear to the breadth shall have been previously
determined. For example, when the proportion is as 6 to 8,
then i J times 6 equals 8, and the proportion is as 1 to i£;
again, when the proportion is as 8 to 10, then ij times 8
equals 10, and in this case the proportion is as i to ij. Let
the latter figure of the ratio i to ij, i to ij, etc., be called
n, or so that the proportion shall be as i to ny then —
To ascertain the thickness of a post to sustain safely a
given weight, when the height is at least ten times the thick-
ness ; the action of the weight coinciding with the axis ; we
have —
Rule X. — Multiply the given weight by the factor of
safety, and divide the product by twice the value of C for
the material of the post, found in Table I., multiplied by ;/,
as above explained ; reserve the quotient, calling it G. Now
multiply 432 times the value of e for the material of the post, .
found in Table III., by the square of the height in feet, and
by the above quotient G ; to the product add the square of
G ; extract the square root of the sum, and to it add th£ value
BREADTH OF POST FOR GIVEN THICKNESS.
95
of G ; then the square root of this sum will be the required
thickness of the post ; or —
Wa
G—-~ TT- -. (12.)
2 C n v '
t = V 1/432 G
(130
Example.— What should be the thickness of a white-pine
rectangular post 20 feet high to sustain safely 30,000 pounds,
the pressure coinciding with the axis, and the proportion
between the thickness and breadth to be as 10 to 12, or as I
to i -2 ? Proceeding according to the rule, we have the pro-
duct of 30000, the given weight, by 6, as a factor of safety,
equals 180000 ; this divided by twice Cx n, or 2 x 6650 x i -2,
(=15960) gives a quotient of 11-278, the value of G. Then,
we have c= -0014, the square of the height equals 400;
therefore, 432 x -0014x400x11.278 = 2728-43. Tothisadd-
ing 127-2, the square of £, we have 2855 63, the square
root of which is 53-438; and this added to G gives 64-716,
the square root of which is 8-045, tne required thickness of
the post. Now, since the thickness is in proportion to the
breadth as i to 1-2, therefore 8-045 x 1-2 = 9-654, the re-
quired width. The post, therefore, may be made 8x9!
inches.
114-.— Breadth of a Rectangular Post.— When the thick-
ness of a post is fixed, and the breadth required ; then, to
ascertain the breadth of a rectangular post to sustain safely
a given weight, the direction of the pressure of which coin-
cides with the axis of the post, we have —
Rule XI. — Divide the height in inches by the given thick-
ness, and multiply the quotient by itself, or take its square ;
multiply this square by the value of e for the material of the
post, found in Table 111. ; to the product add its half, and to
the sum add unity ; multiply this sum by the given weight,
and by the factor of safety ; divide the product by the pro-
duct of the given thickness multiplied by the value of C for
96 CONSTRUCTION.
the material of the post, found in Table I., and the quotient
will be the required breadth ; or—
Example. — What should be the breadth of a spruce post
1 8 feet high and 6 inches thick to sustain safely 25,000
pounds, the pressure coinciding with the axis of the post ?
According to the rule, 216 (= 12 x 18), the height in inches,
divided by 6, the given thickness, gives a quotient of 36, the
square of which is 1296; the value of e for spruce is -00098 ;
this multiplied by 1296, the above square, equals i -27 ; which
increased by -635, its half, amounts to 1-905 ; this increased
by unity, the sum is 2-905 ; which multiplied by the given
weight, and by the factor of safety, gives a product of
435749; and this divided by 6 (the given thickness) times 7850
(the value of C for spruce) = 47 1 oo, gives a quotient of 9 • 2 5 1 6,
the required breadth of the post. The post, therefore, re-
quires to be 6 x 9^ inches.
Observe that when the breadth obtained by the rule is
less than the given thickness, the result shows that the con-
ditions of the case are incompatible with the rule, and that
a new computation must be made ; taking now for the
breadth what was before understood to be the thickness,
and proceeding in this case, by Rule X., to find the thickness.
115. — Resistance to Ten§ion.— In Art. 95 are recorded the
results of experiments made to test the resistance of vari-
ous materials to tensile strain, showing in each case the ca-
pability to such resistance per square inch of sectional area.
The action of materials in resisting a tensile strain is quite
simple ; their resistance is found to be directly as their sec-
tional area. Hence —
116. — The rim it of Weight.— To ascertain the weight or
pressure that may be safely applied to a beam or rod as a
tensile strain, we have —
Rule XII. — Multiply the area of the cross-section of the
beam or rod in inches by the value of 'T, Table II. ; divide
AREA OF BEAM FOR TENSILE STRAIN. 97
the product by the factor of safety, and the quotient will be
the required weight in pounds ; or —
(150
The cross-section here intended is that taken at the small-
est part of the beam or rod. A beam, in framing, is usually
cut with mortices ; the area will probably be smallest at the
severest cutting ; the area used in the rule must be that of the
uncut fibres only.
Example. — The tie-beam of a roof-truss is of white pine,
6 x 10 inches ; the cutting for the foot of the rafter reduces
the uncut area to 40 inches : what amount of horizontal thrust
from the foot of the rafter will this tie-beam safely sustain ?
Here 40 times 12000, the value of T, equals 480000; this
divided by 6, as a factor of safety, gives 80000, the required
weight in pounds.
(17. — Sectional Area. — To ascertain the sectional area of
a beam or rod that will sustain a given weight safely, when
applied as a tensile strain, we have —
Rule XIII. — Multiply the given weight in pounds by the
factor of safety ; divide the product by the value of T, Table
II., and the quotient will be the area required in inches;
or —
A=^. (16.)
This is the area of uncut fibres. If the piece is to be cut
for mortices, or for any other purpose, then for this an
adequate addition is to be made to the result found by the
rule.
Example. — A rafter produces a thrust horizontally of 80,000
pounds ; the tie-beam is to be of oak : what must be the
area of the cross-section of the tie-beam in order to sustain
the rafter safely ? The given weight, 80000, multiplied by
10, as a factor of safety, gives 800000; this divided by 19500,
the value of 7", Table II., the quotient, 41, is the area of uncut
fibres. This should have usually one half of its amount
98 CONSTRUCTION.
added to it as an allowance for cutting; therefore, 41+21
= 62. The tie-beam may be 6 x loj inches.
Another Example. — A tie-rod of American refined
wrought iron is required to sustain safely 36,000 pounds :
what should be its area of cross-section ? Taking 7 as the
factor of safety, 7x36000= 252000; and this divided by
60000, the value of 7", Table II., gives a quotient of 4- 2 inches,
the required area of the rod.
118. — Weight of the Suspending Piece Included.— Pieces
subjected to a tensile strain are frequently suspended verti-
cally. In this case, at the upper end, ,the strain is due not
only to the weight attached at the lower end, but also to the
weight of the rod itself. Usually, in timber, this is small
in comparison with the load, and may be neglected ;
although in very long timbers, and where accuracy is decid-
edly essential, as, also, when the rod is of iron, it may form
a part of the rule. Taking the effect of the weight of the
beam into account, the relation existing between the weights
and the beam requires that the rule for the weight should
be as follows :
Rule XIV. — Divide the value of T for the material of the
beam or rod, Table II., by the factor of safety ; from the
quotient subtract 0-434 times the specific gravity of the ma-
terial in the beam or rod multiplied by the length of the
beam or rod in feet ; multiply the remainder by the area of
cross-section in inches, and the product will be the required
weight in pounds ; or —
W=A -0-434
N. B. — This rule is based upon the condition that the sus-
pending piece be not cut by mortices or in any other way.
Example. — What weight may be safely sustained by a
white-pine rod 4x6 inches, 40 feet long, suspended verti-
cally? For white pine the value of T is 12000; this divid-
ed by 8, as a factor of safety, gives 1 500 ; from which sub-
tracting 0-434 times 0-458 (the specific gravity of white pine,
Table II.) multiplied by 40, the length in feet, the remainder
RESISTANCE TO TRANSVERSE STRAINS. 99
is 1492-049; which multiplied by 24 ( = 4x6, the area of
cross-section) equals 35,761 pounds, the required weight to be
carried. The weight which the rule would give, neglecting
the weight of the rod, would have been 36000; ordinarily,
so slight a difference would be quite unimportant.
119. — Area of Suspending Piece. — To ascertain the area
of a suspended rod to sustain safely a given. weight, when
the^ weight of the suspending piece is regarded, we have —
Rule XV. — Multiply 0-434 times the specific gravity of
the suspending piece by the length in feet ; deduct the pro-
duct from the quotient arising from a division of the value
of T, Table II., by the factor of safety, and with the remain-
der divide the given weight in pounds ; the quotient will be
the required area in inches ; or —
A =
T , ' (18.)
— — o-434/.y
a'
N.B. — This rule is based upon the condition that the rod
be not injured in anywise by cutting.
Example. — What should be the area of a bar of English
cast iron 20 feet long to sustain safely, suspended from its
lower end, a weight of 5000 pounds ? Taking the factor of
safety at 7-0, and the specific gravity also at 7, and the
value of T, Table II., at 17000, we have the product of
0-434 x 7-0 x 20 = 60-76; then 17000 divided by 7 gives
a quotient of 2428-57; from which deducting the above
60-76, there remains 2367-81 ; dividing 5000, the given
weight, by this remainder, we have the quotient, 2-11, which
is the required area in inches.
RESISTANCE TO TRANSVERSE STRAINS.
120. Tran§ver§e Strains: Rupture.— A load placed
upon a beam, laid horizontally or inclined, will bend it, and,
if the weight be proportionally large, will break it. The
power in the material that resists this bending or breaking
is termed the resistance to cross-strains, or transverse strains.
100 CONSTRUCTION.
While in posts or struts the material is compressed or short-
ened, and in ties and suspending pieces it is extended or
lengthened, in beams subjected to cross-strains the material
is both compressed and extended. (See Art. 91.) When the
beam is bent the fibres on the concave side are compressed,
while those on the convex side are extended. The line
where these two portions of the beam meet — that is, the
portion compressed and the portion extended — the hori-
zontal line of juncture, is termed the neutral line or plane.
It is so called because at this line the fibres are neither com-
pressed nor extended, and hence are under no strain what-
ever. The location of this line or plane is not far from the
middle of the depth of the beam, when the strain is not suf-
ficient to injure the elasticity of the material ; but it re-
moves towards the concave or convex side of the beam as
the strain is increased, until, at the period of rupture, its
distance from the top of the beam is in proportion to its dis-
tance from the bottom of the beam as the tensile strength of
the material is to its compressive strength.
121. — Location of Mortices. — In order that the diminution
of the strength of a beam by framing be as small as possible,
all mortices should be located at or near the middle of the
depth. There is a prevalent idea with some, who are aware
that the upper fibres of a beam are compressed when sub-
ject to cross-strains, that it is not injurious to cut these top
fibres, provided that the cutting be for the insertion of an-
other piece of timber — as in the case of gaining the ends of
beams into the side of a girder. They suppose that the piece
filled in will as effectually resist the compression as the part
removed would have done, had it not been taken out. Now,
besides the effect of shrinkage, which of itself is quite suf-
ficient to prevent the proper resistance to the strain, there
is the mechanical difficulty of fitting the joints perfectly
throughout ; and, also, a great loss in the power of resist-
ance, as the material is so much less capable of resistance
when pressed at right angles to the direction of the fibres
than when directly with them, as the results of the experi-
ments in the tables show.
STRENGTH OF BEAMS FOR CROSS-STRAINS. IOI
122. — Transverse Strains : Relation of Weight to Di-
mensions.— The strength of various materials, in their re-
sistance to cross-strains, is given in Table III., Art. 96. The
second column of the table contains the results of experi-
ments made to test their resistance to rupture. In the case
of each material, the figures given and represented by B
indicate the pounds at the middle required to break a unit
of the material, or a piece i inch square and i foot long
between the bearings upon which the piece rests. To be
able to use these indices of strength, in the computation of
the strength of large beams, it is requisite, first, to establish
the relation between the unit of material and the larger
beam. Now, it may be easily comprehended that the strength
of beams will be in proportion to their breadth ; that is,
when the length and depth remain the same, the strength
will be directly as the breadth ; lor it is evident that a beam
2 inches broad will bear twice as much as one which is only
i inch broad, or that one which is 6 inches broad will bear
three times as much as one which is 2 inches broad. This
establishes the relation of the weight to the breadth. With
the depth, however, the relation is different ; the strength is
greater than simply in proportion to the depth. If the
boards cut from a squared piece of timber be piled up in
the order in which they came from the timber, and be loaded
with a heavy weight at the middle, the boards will deflect
or sag much more than they would have done in the timber
before sawing. The greater strength of the material when
in a solid piece of timber is due to the cohesion of the fibres
at the line of separation, by which the several boards, before
sawing, are prevented from sliding upon each other, and
thus the resistance to compression and tension is made to
contribute to the strength. This resistance is found to be
in proportion to the depth. Thus the strength due to the
depth is, first, that which arises from the quantity of the
material (the greater the depth, the more the material),
which is in proportion to the depth ; then, that which en-
sues from the cohesion of the fibres in such a manner as to
prevent sliding ; this is also as the depth. Combining the
two, we have, as the total result, the resistance in proportion
102 CONSTRUCTION.
to the square of the depth. The relation between the
weight and the length is such that the longer the beam is,
the less it will resist ; a beam which is 20 feet long will sus-
tain only half as much as one which is 10 feet long ; the
breadth and depth each being the same in the two beams.
From this it results that the resistance is inversely in pro-
portion to the length. To obtain, therefore, the relation
between the strength of the unit of material and that of a
larger beam, we have these facts, namely : the strength of
the unit is the value of B, as recorded in Table III. ; and
the strength of the larger beam, represented . by W, the
weight required to break it, is the product of the breadth
into the square of the depth, divided by the length ; or,
while for the unit we have the ratio —
B\ i,
we have for the larger beam the ratio —
Therefore, putting these ratios in an expressed proportion,
we have —
From which (the product of the means equalling the pro-
duct of the extremes ; see Art. 373) we have—
w =
In which W represents the pounds required to break a
beam, when acting at the middle between the two supports
upon which the beam is laid ; of which beam b represents
the breadth and d the depth, both in inches, and / the length
in feet between the supports ; and B is from Table III., and
represents the pounds required to break a unit of material
like that contained in the larger beam.
fuinvERsm
LIMIT OF WEIGHT AT MIDDLE. ?&&,>'
123.— Safe Weight: Load at Middle.— The relation
established, in the last article, between the weight and the
dimensions is that which exists at the moment of rupture.
The rule (19.) derived therefrom is not, therefore, directly
practicable for computing the dimensions of beams for
buildings. From it, however, one may readily be deduced
which shall be practicable. In the fifth column of Table III.
are given the least values of a, the factor of safety, explained
in Art. 96. Now, if in place of By the symbol for the break-
ing weight, the quotient of B divided by a be substituted,
then the rule at once becomes practicable ; the results now
being in consonance with the requirements for materials
used in buildings. Thus, with this modification, we have —
Therefore, to ascertain the weight which a beam may be
safely loaded with at the centre, we have—
Rule XVI.— Multiply the value of B, Table HI., for the
kind of material in the beam by the breadth and by the
square of the depth of the beam in inches ; divide the pro-
duct by the product of the factor of safety into the length
of the beam between bearings in feet, and the quotient will
be the weight in pounds that the beam will safely sustain
at the middle of its length.
Example. — What weight in pounds can be suspended
safely from the middle of a Georgia-pine beam 4x 10 inches,
and 20 feet long between the bearings ? For Georgia pine the
value of B, in Table III., is 850, and the least value of a is
1-84. For reasons given in Art. 96, let a be taken as high
as 4; then, in this case, the value of b is 4, and that of d is
10,' while that of/ is 20. Therefore, proceeding by the rule,
850 x 4 x io2 = 340000 ; this divided by 4 x 20 (— 80) gives
a quotient of 4250 pounds, the required weight.
Observe that, had the value of a been taken at 3, instead
of 4, the result by the rule would have been a load of 5667
pounds, instead of 4250, and the larger amount would be
none too much for a safe load upon such a beam ; although,
104 CONSTRUCTION.
with it, the deflection would be one third greater than with
the lesser load. The value of a should always be assigned
higher than the figures of the table, which show it at its
least value ; but just how much higher must depend upon
the firmness required and the conditions of each particular
case.
124. — Breadth of Beam willi Safe Load. — By a simple
transposition of the factors in equation (20.), we obtain —
a rule for the breadth of the beam.
Therefore, to ascertain what should be the breadth of a
beam of given depth and length to safely sustain at the
middle a given weight, we have —
Rule XVII. — Multiply the given weight in pounds by
the factor of safety, and by the length in feet, and divide the
product by the square of the depth multiplied by the value
of B for the material in the beam, in Table III. ; the quotient
will be the required breadth.
Example. — What should be the breadth of a white-pine
beam 8 inches deep and 10 feet long between bearings to
sustain safely 2400 pounds at the middle ? For white pine
the value of B, in Table III., is 500. Taking the value of a
at 4, and proceeding by the rule, we have 2400 x4x 10 =
96000 ; this divided by (8a x 500 =) 32000 gives a quotient
of 3, the required breadth of the beam.
125. — Depth of Beam with Safe Load. — A transposition
of the factors in equation (21.), and marking it for extraction
of the square root, gives —
a rule for the depth of a beam. Therefore, to ascertain what
should be the depth of a beam of given breadth and length
to safely sustain a given weight at the middle, we have —
WEIGHT AT ANY POINT.
105
Rule XVIIL— Multiply the given weight by the factor of
safety, and by the length in feet ; divide the product by the
product of the breadth into the value of B for the kind of
wood, Table III. ; then the square root of the quotient will
be the required depth.
Example. — What should be the depth of a spruce beam
5 inches broad and lofeet long between bearings to sustain
safely, at middle, 4500 pounds ? The value of B from the table
is 550; taking a at 4, and proceeding by the rule, we have
4500 x 4 x 15 = 270000; this divided by (550 x 5 =) 2750
gives a quotient of 98-18, the square root of which is 9-909,
the required depth of the beam. The beam should be 5 x 10
inches.
126. — Safe Load at any Point. — When the load is at the
middle of a beam it exerts the greatest possible strain ; at
any other point the strain would be less. The strain de-
creases gradually as it approaches one of the bearings, and
when arrived at the bearing its effect upon the beam as a
cross-strain is zero. The effect of a weight upon a beam is
in proportion to its distance from one of the bearings, mul-
tiplied by the portion of the load borne by that bearing.
The load upon a beam is divided upon the two bearings,
as shown at Art. 88. The weight which is required to rup-
ture a beam is in proportion to the breadth and square of
the depth, b d*, as before shown, and also in proportion to
the length divided by 4 times the rectangle of the two parts
into which the load divides the length, or (see Fig. 35).
4 MI 11
This, when the load is at the middle, may be put as
= i a result coinciding with the relation before
4x|./xi/ /
given in Art. 122, viz. : "The resistance is inversely in pro-
portion to the length." The total resistance, therefore, put-
b d* I
ting the two statements together, is in proportion to -
A /// /£
7727
There are, therefore, these two ratios, viz., W \ - - and
4 m i*
B : I, from which we have the proportion—
106 CONSTRUCTION.
4 m n
from which we have —
4 ;« n
(23.}
r>
This is the relation at the point of rupture, and when — is
used instead of B, the expression gives the safe weight.
Therefore —
(24.)
4 a m n
is an expression for the safe weight. Now, to ascertain the
weight which may be safely borne by a beam at any point
in its length, we have —
Rule XIX. — Multiply the breadth by the square of the
depth, by the length in feet, and by the value of B for the
material of the beam, in Table III. ; divide the product by
the product of four times the factor of safety into the rec-
tangle of the two parts into which the centre of gravity of
the weight divides the beam, and the quotient will be the
required weight in pounds.
Example. — What weight may be safely sustained at 3 feet
from one end of a Georgia-pine beam which is 4 x 10 inches,
and 20 feet long? The value of B for Georgia pine, in
Table III., is 850 ; therefore, by the rule, 4 x ioa x 20 x 850 =
6800000. Taking the factor of safety at 4, we have
4x4x3x17=816. Using this as a divisor with which to
divide the former product, we have as a quotient 8333
pounds, the required weight.
127. — Breadth or Depth: Load at any Point. — By a
proper transposition of the factors of (24.) we obtain —
, ,a 4 W a in 11
an expression showing- the product of the breadth into the
square of the depth ; hence, to ascertain the breadth or
DISTRIBUTED WEIGHT. IO/
depth of a beam to sustain safely a given weight located at
any point on the beam, we have —
Rule XX. — Multiply four times the given weight by the
factor of safety, and by the rectangle of the two parts into
which the load divides the length ; divide the product by
the product of the length into the value of B for the mate-
rial of the beam, found in Table III., and the quotient will be
equal to the product of the breadth into the square of the
depth. Now, to obtain the breadth, divide this product by
the square of the depth, and the quotient will be the required
breadth. But if, instead of the breadth, the depth be de-
sired, divide the said product by the breadth ; then the
square root of the quotient will be the required depth.
Example. — What should be the breadth (the depth being
8) of a white-pine beam 12 feet long to safely sustain 3500
pounds at 3 feet from one end ? Also, what should oe its
depth when the breadth is 3 inches? By the rule, taking
the factor of safety at 4, 4 x 3500 x 4 x 3 x 9 = 1512000.
The value of B for white pine, in Table III., is 500 ; there-
fore, 500 x 12 = 6000; with this as divisor, dividing 1512000,
the quotient is 252. Now, to obtain the breadth when the
depth is 8, 252 divided by (8 x 8 =) 64 gives a quotient of
3-9375, the required breadth ; or the beam may be, say, 4 x 8.
Again, when the breadth is 3 inches, we have for the quotient
of 252 divided by 3 = 84, and the square root of 84 is 9- 165,
or 9^ inches. For this case, therefore, the beam should be,
say, 3 x 9 J inches.
128. — Weight Uniformly Distributed.— When the load is
spread out uniformly over the length of a beam, the beam
will require just twice the weight to break it that would be
required if the weight were concentrated at the centre.
Therefore, we have W — —, where U represents the dis-
tributed load. Substituting this value of W in equation
(20.), we have —
U__Bbd\
2 ~~ ' a I '
U=2-*** (26.) -
a I
IOS CONSTRUCTION.
Therefore, to ascertain the weight which may be safely sus-
tained, when uniformly distributed over the length of a
beam, we have —
Rule XXI. — Multiply twice the breadth by the square of
the depth, and by the value of B for the material of the
beam, in Table III., and divide the product by the product
of the length in feet by the factor of safety, and the quotient
will be the required weight in pounds.
Example. — What weight uniformly distributed may be
safely sustained upon a hemlock beam 4x9 inches, and 20
feet long? The value of B for hemlock, in Table III., is
450 ; therefore, by the rule, 2 x 4 x 9' x 450 = 291600. Tak-
ing the factor of safety at 4, we have 4 x 20 — 80, the pro-
duct by which the former product is to be divided. This
division produces a quotient of 3645, the required weight.
129.— Breadth or Depth : Load Uniformly Distributed.—
By a proper transposition of factors in (26.), we obtain —
an expression giving the value of the breadth into the square
of the depth. From this, therefore, to ascertain the breadth
or the depth of a beam to sustain safely a given weight uni-
formly distributed over the length of a beam, we have —
Rule XXII. — Multiply the given weight by the factor of
safety, and by the length ; divide the product by the pro-
duct of twice the value of B for the material of the beam,
in Table III., and the quotient will be equal to the breadth
into the square of the depth. Now, to find the breadth,
divide the said quotient by the square of the depth ; but if,
instead of the breadth, the depth be required, then divide
said quotient by the breadth, and the square root of this
quotient will be the required depth.
Example. — What should be the size of a white-pine beam 20
feet long to sustain safely 10,000 pounds uniformly distributed
over its length ? The value of B for white pine, in Table III.,
is 500. Let the factor of safety be taken at 4. Then, by the
rule, loooo x 4 x 20 = 800000 ; this divided by (2 x 500 =)
WEIGHT PER BEAM IN FLOORS. 109
1000 gives a quotient of 800. Now, if the depth be fixed at
12, then the said quotient, 800, divided by (12 x 12=) 144
gives 5-5-, the required breadth of beam ; and the beam may
be, say, 5! x 12. Again, if the breadth is fixed, say, at 6, and
the depth is required, then the said quotient, 8co, divided by
6 gives 133^, the square root of which, 1 1 - 55, is the required
depth. The beam in this case should therefore be, say,
6 x I if inches.
130. — [Load per Foot Superficial. — When several^beams
are laid in a tier, placed at equal distances apart, as in a tier
of floor-beams, it is desirable to know what should be their
size in order to sustain a load equally distributed over the
floor.
If the distance apart at which they are placed, measured
from the centres of the beams, be multiplied by the length
of the beams between bearings, the product will equal the
area of the floor sustained by one beam ; and if this area be
multiplied by the weight upon a superficial foot of the floor,
the product will equal the total load uniformly distributed
over the length of the beam ; or, if c be put to represent the
distance apart between the centres of the beams in feet, and
/ represent the length in feet of the beam between bearings,
and/ equal the pounds per superficial foot on the floor,
then the product of these, or c f I, will represent the uni-
formly distributed load on a beam ; but this load was before
represented by U (Art. 128); therefore, we have cfl= U,
and they may be substituted for it in (26.) and (27.). Thus
we have —
b d* = cflal
or —
(28.)
Therefore, to ascertain the size of floor-beams to sustain
safely a given load per superficial foot, we have—
Rule XXIII.— Multiply the given weight per superficial
foot by the factor of safety, by the distance between the
HO CONSTRUCTION.
centres of the beams in feet, and by the square of the length
in feet; divide the product by twice the value of B for the
material of the beams, in Table III., and the quotient will be
equal to the breadth into the square of the depth. Now, to
obtain the breadth, divide said quotient by the square of the
depth, and this quotient will be the required breadth. But
if, instead of the breadth, the depth be required, divide the
aforesaid quotient by the breadth ; then the square root of
this quotient will be the required depth.
Example. — What should be the size of white-pine floor-
beams 20 feet long, placed i# inches from centres, to sustain
safely 90 pounds per superficial foot, including the weight
of the materials of construction — the beams, flooring, plas-
tering, etc. ? The value of B for white pine is 500 ; the
factor of safety may be put at 5. Then, by the rule, we
have 90 x 5 x -ff x 2O2 = 240000. This divided by (2 x 500
=) 1000 gives 240. Now, for the breadth, if the depth be
fixed at 9 inches, then 240 divided by (9* = ) 8 1 gives a
quotient of 2-963. The beams therefore should be, say,
3x9. But if the breadth be fixed, say, at 2-5 inches, then
240 divided by 2-5 gives a quotient of 96, the square root of
which is 9-8 nearly. The beams in this case would require
therefore to be, say, 2^ x 10 inches.
N. B. — It is well to observe that the question decided
by Rule XXII. is simply that of strcngtli only. Floor-beams
computed by it will be quite safe against rupture, but they
will in most cases deflect much more than would be consist-
ent with their good appearance. Floor-beams should be
computed by the rules which include the effect of deflection.
(See Art. 152.)
131. — Levers: Load at One End.— The beams so far con-
sidered as being exposed to transverse strains have been
supposed to be supported at each end. When a piece is
held firmly at one end only, and loaded at the other, it is
termed a lever ; and the load which a piece so held and
loaded will sustain is equal to one fourth that which the
same piece would sustain if it were supported at each end
and loaded at the middle. Or, the strain in a beam sup-
LEVERS TO SUSTAIN GIVEN WEIGHTS. Ill
ported at each end caused by a given weight located at the
middle is equal to that in a lever of the same breadth and
depth, when the length of the latter is equal to on6 half that
of the beam, and the load at its end is equal to one half of
that at the middle of the beam. Or, when P represents the
load at the end of the lever, and n its length, then W—2P,
and l—2n. Substituting these values of W and / in equa-
tion (20.), we have —
4
2an
from which —
p-Bbd*
T^P
Hence, to ascertain the weight which may be safely sus-
tained at the end of a lever, we have —
Rule XXIV.— Multiply the breadth of the lever by the
square of its depth, and by the value of B for the material
of the lever, in Table III. ; divide the product by the pro-
duct of four times the length in feet into the factor of safety,
and the quotient will be the required weight in pounds.
Example. — What weight can be safely sustained at the
end of a maple lever of which the breadth is 2 inches, the
depth is 4 inches, and the length is 6 feet ? The value of B
for maple, in Table III., is uoo; therefore, by the rule,
2 x42 x i TOO = 35200. And, taking the factor of safety at 5,
4x5x6= 1 20, and 35200 divided by 120 gives a quotient of
293 '33> or 293^- pounds.
N. B. — When a lever is loaded with a weight uniformly
distributed over its length, it will sustain just twice the load
which can be sustained at the end.
132. — Levers: Breadth or Depth. — By a proper trans-
position of the factors in (29.), we obtain—
L (30.)
Z>
Hence, to ascertain the breadth or depth of a lever to sus-
tain safely a given weight, we have —
II2 CONSTRUCTION.
Rule XXV. — Multiply four times the given weight by
the length of the lever, and by the factor of safety ; divide
the product by the value of B for the material of the lever,
in Table III., and the quotient will be equal to the breadth
multiplied by the square of the depth. Now, if the breadth
be required, divide said quotient by the square of the depth,
and this quotient will be the required breadth ; but if,
instead of the breadth, the depth be required, divide the
said quotient by the breadth ; then the square root of this
quotient will be the required depth.
Example. — What should be the size of a cherry lever 5
feet long to sustain safely 250 pounds at its end? Proceed-
ing by the rule, taking the factor of safety at 5, we have
4x250x5x5 = 25000. The value of B for cherry, in Table
III., is 650 ; and 25000 divided by 650 gives a quotient of
38-46. Now, if the depth be fixed at 4, then 38-46 divided
by (4x4 =) 16 gives a quotient of 2-4, the required breadth.
But if the breadth be fixed at 2, then 38-46 divided by 2
gives a quotient of 19-23, the square root of which is 4-38,
the required depth. Therefore, the lever maybe 2-4x4,
or 2 x 4-f- inches.
133. — Deflection: Relation to Weight. — When a load is
placed upon a beam supported at each end, the beam bends
more or less ; the distance that the beam descends under
the operation of the load, measured at the middle of .its
length, is termed its deflection. In an investigation of the
laws of deflection it has been demonstrated, and experiments
have confirmed it, that while the elasticity of the material
remains uninjured by the pressure, or is injured in but a
small degree, the amount of deflection is directly in propor-
tion to the weight producing it ; for example, if 1000 pounds
laid upon a beam is found to cause it to deflect or descend at
the middle a quarter of an inch, then 2000 pounds will cause
it to deflect half an inch, 3000 pounds will deflect it three
fourths of an inch, and so on.
134. — Deflection : Relation to Dimensions. — In Table
III. are recorded the results of experiments made to test the
THE LAW OF DEFLECTION. 113
resistance of the materials named to deflection. The fig-
ures in the third column designated by the letter F (for flex-
ure) show the number of pounds required to deflect a unit
of material one inch. This is an extreme state of the case,
for in most kinds of material this amount of depression
would exceed the limits of elasticity ; and hence the rule
would here fail to give the correct relation as between the
dimensions and pressure. For the law of deflection as above
stated (the deflections being in proportion to the weights)
is true only while the depressions are small in comparison
with the length. Nothing useful is, therefore, derived from
this position of the question, except to give an idea of the
nature of the quantity represented by the constant F\ it
being in reality an index of the stiffness of the kind of mate-
rial used in comparing one material with another. Whatever
be the dimensions of the beam, F will always be the same
quantity for the same material ; but among various materials
/''will vary according to the flexibility or stiffness of each
particular material. For example, F will be much greater
for iron than for wood ; and again, among the various kinds
of wood, it will be larger for the stiff woods than for those
that are flexible. The value of F, therefore, is the weight
which would deflect the unit of material one inch, upon the
supposition that the deflections, from zero to the depth of
one inch, continue regularly in proportion to the increments
of weight producing the deflections, or, for each deflection —
F : I : : W : <?,
from which we have —
in which cJ represents the deflection in inches corresponding
to Wt the weight producing it. This is for the unit of ma-
terial. For beams of larger dimensions, investigations have
shown (Transverse Strains, Chapters XIII. and XIV.) that
the power of a beam to resist deflection by a weight at mid-
dle is in proportion to its breadth and the cube of its depth,
and it is inversely in proportion to the cube of the length ;
1 14 CONSTRUCTION.
or, when the resistance of the unit of material is measured,
as above, by — , we have the relation between it and a
larger beam of —
Putting this ratio in a proportion with that of the unit of
material, we have —
. ..; . . ; ";i>: ' :: T 'b-r-- :
which gives —
W _Fbd*
6 ~ T '
from which we have —
W=F-^-. (31.)
135. — Deflection : Weight when at MidklQc. — In equation
(31.) we have a rule by which to ascertain what weight is
required to deflect a given beam to a given depth of deflec-
tion ; this, in words at length, is —
Rule XXVI.— Multiply the breadth of the beam by the
cube of its depth, and by the given deflection, all in inches,
and by the value of .Ffor the material of the beam, in Table
III.; divide the product by the cube of the length in feet,
and the quotient will be the required weight in pounds.
Example. — What weight is required at the middle of a
4x12 inch Georgia-pine beam 20 feet long to deflect it
three quarters of an inch ? The value of F for Georgia
pine, in Table III., is 5900; therefore, by the rule, we have
4 x i23 x 0-75 x 5900 = 30585600, which divided by (20x20
x 20 =) 8000 gives a quotient of 3823-2, the required weight
in pounds.
136.— Deflection: Breadth or Depth, Weight at middle.
—By a transposition of equation (31.), we obtain—
SIZE FOR A GIVEN DEFLECTION. 115
a rule by which may be found the breadth or depth of a
beam, with a given load at middle and with a given deflec-
tion ; this, in words at length, is —
Rule XXVII. — Multiply the given load by the cube of
the length in feet, and divide the product by the product of
the deflection into the value of F for the material of the
beam, in Table III. ; then the quotient will be equal to the
breadth of the beam multiplied by the cube of its depth,
both in inches.
Now, to obtain the breadth, divide the said quotient by
the cube of the depth, and this quotient will be the required
breadth. But if, instead of the breadth, the depth be re-
quired, then divide the said quotient by the breadth, and
the cube root of this quotient will be the required depth.
But if neither breadth nor depth be previously fixed, but it
be required that they bear a certain proportion to each
other ; such that d : b : : i : r, r being a decimal, then b = rd,
and b d* — r d* ; then, to find the depth, divide the aforesaid
quotient by the decimal r, and the fourth root (or the square
root of the square root) will be the required depth, and this
multiplied by the decimal r will give the breadth.
Example. — What should be the size of a spruce beam 20
feet long between bearings, sustaining 2000 pounds at the
middle, with a deflection of one inch ? By the rule, the
weight into the cube of the length is 2000 x 8000 = 16000000.
The value of Ffor spruce, in Table III., is 3500; this by the
deflection = i gives 3500, which used as a divisor in divid-
ing the above 16000000 gives a quotient of 4571 -43. Now,
if the breadth be required, the depth being fixed, say, at 10,
then 4571-43 divided by (lox lox 10 =) 1000 gives 4-57, the
required breadth. The beam should be, say, 4$ by 10 inches.
But if the depth be required, the breadth being fixed, say, at
4, then 4571-43 divided by 4 gives 1142-86, the cube root
of which is 10-46; so in this case, therefore, the beam is
required to be 4 x io£ inches. Again, if the breadth is to
bear a certain proportion to the depth, or that the ratio be-
tween them is to be, say, 0-6 to i, then let r = 0-6, and then
457i.43 = o-6^4, and dividing by 0-6, we have 7619-05
= d\ This equals d*xd*\ therefore the square root of 7619
1 16 CONSTRUCTION.
is 87-29, and the square root of this is 9-343, the required
depth in inches. Now 9-343x0-6 equals the breadth, or
9.343x0-6=5-6; therefore the beam is required to be
5 -6 x 9- 34 inches, or, say, 5f x 9^ inches.
137.— Deflection : when Weight i§ at Middle. — By a trans-
position of the factors in (32.), we obtain —
a rule by which the deflection of any given beam may be as-
certained, and which, in words at length, is —
Rule XXVIIL— Multiply the given weight by the cube
of the length in feet ; divide the product by the product of
the breadth into the cube of the depth in inches, multiplied
by the value of Fior the material of the beam, in Table III.,
and the quotient will be the required deflection in inches!
Example. — To what depth will 1000 pounds deflect a
3x10 inch white-pine beam 20 feet long, the weight being
at the middle of the beam ? By the rule, we have 1000 x 2o3
= 8000000; then, since the value of F for white pine, in
Table III., is 2900, we have 3 x io3 x 2900 = 8700000 ; using
this product as a divisor and by it dividing the former pro-
duct, we obtain a quotient of 0.9195, the required deflection
in inches.
138. — Deflection: Load . Uniformly Distributed. — In two
beams of equal capacity, suppose the one loaded at the
middle, and the other with its load uniformly distributed
over its length, and so loaded that the deflection in one beam
shall equal that in the other ; then the weight at the middle
of the former beam will be equal to five eighths of that on
the latter. This proportion between the two has been de-
monstrated by writers on the strength of materials. (See p.
484, Mechanics of Eng. and Arch., by Prof. Mosely, Am. eel. by
Prof. Mahan, 1856.) Hence, when £/is put to represent the
uniformly distributed load, we have—
DEFLECTION FOR LOAD EQUALLY DISTRIBUTED. 1 17
or, when an equally distributed load deflects a beam to a
certain depth, five eighths of that load, if concentrated at
the middle, would cause an equal deflection. This value of
W may therefore be substituted for it in equation (31.), and
give—
from which we obtain —
i-
u= -- JT -- > (34-)
a rule for a uniformly distributed load.
139. — Deflection: Weight when Uniformly Distributed.
— In equation (34.) we have a rule by which we may ascertain
what weight is required to deflect to a given depth any
given beam. This, in words at length, is —
Rule XXIX. — Multiply 1-6 times the deflection by the
breadth of the beam, and by the cube of its depth, all in
inches, and by the value of Ffor the material of the beam,
in Table III. ; divide the product by the cube of the length
in feet, and the quotient will be the required weight in
pounds.
Example. — What weight, uniformly distributed over the
length of a spruce beam, will be required to deflect it to the
depth ot 0-5 ot an inch, the beam being 3 x 10 inches and 10
feet long? The value of F ior spruce, in Table III., is 3500.
Therefore, by the rule, we have j -6x0-5 x 3 x io3 x 3500 =
8400000, and this divided by (10x10x10=) 1000 gives
8400, the required weight in pounds.
(40, __ Deflection: Breadth or Depth, Load Uniformly
Distributed. — By transposition of the factors in equation
(54.), we obtain —
a rule for the dimensions, which, in words at length, is —
Il8 CONSTRUCTION.
Rule XXX. — Multiply the given weight by the cube of
the length of the beam ; divide the product by i -6 times the
given deflection in inches, multiplied by the value of F for
the material of the beam, in Table III., and the quotient will
equal the breadth into the cube of the depth. Now, to ob-
tain the breadth, divide this quotient by the cube of the depth,
and the resulting quotient will be the required breadth in
inches. But if, instead of the breadth, the depth be required,
then divide the aforesaid quotient by the breadth, and the
cube root of the resulting quotient will be the required depth
in inches. Again, if neither breadth nor depth be previously
determined, but to be in proportion to each other at a given
ratio, as r to i, r being a decimal fixed at pleasure, then di-
vide the aforesaid quotient by the value of r, and take the
square root of the quotient; then the square root of this
square root will be the required depth in inches. The breadth
will equal the depth multiplied by the value of the deci-
mal r.
Example. — What should be the size of a locust beam 10
feet long which is to be loaded with 6000 pounds equally
distributed over the length, and with which the beam is to
be deflected £ of an inch ? The value of F for locust, in
Table 1 1 1., is 5050. By the rule, we have 6000 x(io x 10 x 10 =)
1000 = 6000000, which is to be divided by (i -6xo- 75 x 5050 =)
6060, giving a quotient of 990-1. Now, if the depth be, say,
6 inches, then 990- 1 divided by (6x6x6—) 216 gives a quo-
tient of 4-584, the required breadth in inches, say 4^. But
if the breadth be assumed at 4 inches, then 990- 1 divided by
4.gives a quotient of 247- 5 2 5, the cube root of which is 6-279,
the required depth in inches, or, say, 6J. And, again, if the
ratio between the breadth and depth be as o- 7 to i, then 990- 1
divided by 0-7 gives a quotient of 1414-43, the square root
of which is 37-609, of which the square root is 6-1326, the
required depth in inches, or, say, 6J- ; and then 6-1326x0-7 =
4-293, the required breadth in inches; or, the beam shoujd
be 4T\ x 6J- inches.
14-1. — Deflection : when Weight is Uniformly Distributed.
—By a transposition of the factors of equation (35.), we ob-
tain —
DEFLECTION OF LEVERS AND BEAMS. 119
a result nearly the same as that in equation (33.), which is a
rule for deflection by a weight at middle, and which by
slight modifications may be used for deflection by an equally
distributed load. Thus by—
Rule XXXI.— Proceed as directed in Rule XXVIII. (Art.
137), using the equally distributed weight instead of a con-
centrated weight, and then divide the result there obtained
for deflection by I -6 ; then the quotient will be the required
deflection in inches.
Example. — Taking the example given under Rule
XX VI 1 1., \uArt. 137, and assuming that the 1000 pounds load
with which the beam is loaded be equally distributed, then
0-9195, the result for deflection as there found, divided by i -6,
as by the above rule, gives 0-5747, the required deflection.
This result is just five eighths of 0-9195, the deflection by the
load at middle.
N.B. — The deflection by a uniformly distributed load is
just five eighths of that produced by the same load when
concentrated at the middle of the beam; therefore, five
eighths of the deflection obtained by Rule XXVIII. will be
the deflection of the same beam when the same weight is
uniformly distributed.
142. — Deflection of Levers. — The deflection of a lever is
the same as that of a beam of the same breadth and depth,
but of twice the length, and loaded at the middle with a load
equal to twice that which is at the end of the lever. There-
fore, if P represents the weight at the end of a lever, and n
the length of the lever in feet, then 2 P= W smd 2 n = t, and
if these values of Wand /be substituted for those in equa-
tion (33.), we obtain —
2 P x 2 n 3
which reduces to —
-.„,,., (37-)
Fbd"
120 CONSTRUCTION.
a result 16 times that in equation (33.), which is the deflection
in a beam. Therefore, when a beam and a lever equal in
sectional area and in length be loaded by equal weights, the
one at the middle, the other at one end, the deflection of the
lever will be 16 times that of the beam. This proportion is
based upon the condition that neither the beam nor the lever
shall be deflected beyond the limits of elasticity.
14-3. — Deflection of a Lever: Load at End. — Equation
(37.), in words at length, is —
Ride XXXII. — Multiply 16 times the given weight by
the cube of the length in feet ; divide the product by the
product of the breadth into the cube of the depth multiplied
by the value of .Ffor the material of the lever, in Table III.,
and the quotient will be the required deflection.
Example. — What would be the deflection of a bar of
American wrought iron one inch broad, two inches deep,
loaded with 150 pounds at a point 5 feet distant from the
wall in which the bar is imbedded ? The value of F for
American wrought iron, in Table III., is 62000. Therefore,
by the rule, 16 x 150 x 5" = 300000. This divided by
(i x 23 x 62000 =) 496000 gives 0-6048, the required deflec-
tion — nearly f of an inch.
(4-4-. — Deflection of a Lever: Weight when at End. — By
a transposition of the factors in equation (37.), we obtain —
This result is equal to one sixteenth of that shown in equa-
tion (31.), a rule for the weight at the middle. Therefore,
for—
Rule XXXIII.— Proceed as directed in Rule XXVII.;
divide the quotient there obtained by 16, and the resulting
quotient will be the required weight in pounds.
Example. — What weight is required at the end of a 4 x 12
inch Georgia-pine lever 20 feet long to deflect it three
quarters of an inch? Proceeding by Rule XXVII., we ob-
tain a quotient of 3823-2; this divided by 16 gives
say 239, the required weight in pounds.
DEFLECTION OF LEVERS WITH UNIFORM LOAD. 121
145. — Deflection of a Lever : Breadth or Depth, Load
at End. — A transposition of the factors of equation (38.)
gives—
, 73 i6Pn*
a rule by which to obtain the sectional area of the lever.
By comparison with equation (32.) it is seen that the result
in (39.) is 16 times that found by (32.). Therefore, the dimen-
sions for a lever loaded at the end may be found by—
Rule XXXIV. — Multiply by 16 the first quotient found
by Rule XXVII., and then proceed as farther directed in
Rule XXVII., using the product of 16 times the quotient,
instead of the said quotient.
Example. — What should be the size of a spruce lever 20
feet long, between weight and wall, to sustain 2000 pounds
at the end with a deflection of I inch? Proceeding by Rule
XXVII., we obtain a first quotient of 4571-43. By Rule
XXXIV., 4571-43 x 16 — 73144-88. Now, if the depth be
fixed, say, at 20, then 73144-88 divided by (20 x 20 x 20 =)
8000 gives 9- 143, the required breadth. But to obtain the
depth, fixing the breadth, say, at 9, we have for 73144-88 di-
vided by 9 = 8127-21, the cube root of which is- 20- 1055, the
required depth. Again, if the breadth and depth are to be
in proportion, say, as 0-7 to i-o, then 73144-88 divided by
0-7 gives 104492-7, the square root of which is 323-254, of
which the square root is 17-98, the required depth in inches ;
and 17-98 x 0-7 = 12-586, the required breadth in inches.
The lever, therefore, should be, say, I2f x 18 inches.
146. — Deflection of Levers: Weight Uniformly Distrib-
uted.— A comparison of the effects of loads upon levers
shows (Transverse Strains, Art. 347) that the deflection by a
uniformly distributed load is equal to that which would be
produced by three eighths of that load if suspended from
the end of the lever. Or, P — f U. Substituting this value
of P, in equation (37.), gives—
16 x | Un*
Fb<T* '
which reduces to —
122 CONSTRUCTION.
a rule for the deflection of levers loaded with an equally dis-
tributed load.
(47. _ Deflection of Severs with Uniformly Distributed
Load. — The deflection shown in equation .(40.) is just six
times that shown in equation (33.). The result by (33.) mul-
tiplied by 6 will equal the result by (40.); therefore, we
have —
Rule XXXV.— Proceed as directed in Rule XXVIII. ;
the result thereby obtained multiplied by 6 will give the
required deflection.
Example. — To what depth will 500 pounds deflect a 3 x 10
inch white-pine lever 10 feet long, the weight uniformly
distributed over the lever? Here, by Rule XXVIII. , we
obtain the result 0-05747 ; this multiplied by 6 gives 0-3448,
the required deflection.
(4-8. — Deflection of Levers : Weight when Uniformly
Distributed. — By a transposition of factors in (40.), we ob-
tain —
This is equal to one sixth that of equation (31.) ; therefore,
we have —
Rule XXX VI. — Proceed as directed in Rule XXVI.;
the quotient thereby obtained divide by 6, and the quotient
thus obtained will be the required weight.
Example. — What weight will be required to deflect a
4x5 inch spruce lever I inch, the weight uniformly dis-
tributed over its length ? Proceeding as directed in Rule
XXVI., the result thereby obtained is 1750; this divided by
6 gives 29 if, the required weight in pounds.
14-9. — Deflection of Severs : ISreadth or Depth, Load
Uniformly Distributed. — A transposition of factors in equa-
tion (41.) gives —
SIMPLICITY IN CONSTRUCTION. 123
This result is just six times that of equation (32.); we, there-
fore, have —
Rule XXXVII.— Proceed as directed in Rule XXVII. ;
multiply the first quotient thereby obtained by 6 ; then in
the subsequent directions use this multiplied quotient in-
stead of the said first quotient, to obtain the required breadth
and depth.
Example. — What should be the size of a spruce lever 10
feet long, sustaining 2666| pounds, uniformly distributed
over its length, with a deflection of I inch ? Proceeding
by Rule XXVII., the first quotient obtained is 761-905;
this multiplied by 6 gives 4571-43, the multiplied quotient
which is to be used in place of the said first quotient. Now,
to obtain the breadth, the depth being fixed, say, at 10 ;
4571 -43 divided by (cube of 10 — ) 1000, the quotient, 4-57, is
the required breadth. But if the breadth be fixed, say, at
4, then, to obtain the depth, 4571-43 divided by 4 gives
1142-86, the cube root of which is 10-46, the required depth.
Again, if the breadth and depth are to be in proportion, say,
as 0-6 to i -o, then 4571 -43 divided by 0-6 gives 7619-05, the
square root of which is 87-27, of which the square root is
9-343, the required depth in inches ; and 9-343 x °'6 equals
5-6, the required breadth in inches; or, the lever may be,
say, 5f x 9§- inches.
CONSTRUCTION IN GENERAL.
150. — Construction: Object Clearly Defined. — In the
various parts of timber construction, known as floors, par-
titions, roofs, bridges, etc., each has a specific object, and in
all designs for such constructions this object should be kept
clearly in view, the various parts being so disposed as to
serve the design with the least quantity of material. The
simplest form is the best, not only because it is the most
economical, but for many other reasons. The great number
of joints, in a complex design, render the construction liable
to derangement by multiplied compressions, shrinkage, and,
in consequence, highly increased oblique strains ; by which
its stability and durability are greatly lessened.
I24
CONSTRUCTION.
FLOORS.
151. — Floor§ Described. — Floors are most generally con,
structed single; that is, simply a series of parallel beams, each
FIG. 39.
spanning the width of the building, as seen at Fig. 39* Oc-
FIG. 40.
casionally floors are constructed double, as at Fig. 40 ; and
sometimes framed, as at Fig. 41 ; but these methods are
RULES APPLIED TO FLOORS.
I25
seldom practised, inasmuch as either of these requires more
timber than the single floor. Where lathing and plastering
is attached to the floor-beams to form a ceiling below, the
springing of the beams, by customary use, is liable to crack
the plastering. To obviate this in good dwellings, the double
and framed floors have been resorted to, but more in former
times than now, as the cross-furring (a series of narrow strips
of board or plank nailed transversely to the underside of
UNIVERSITY
FIG. 41.
the beams to receive the lathing for the plastering) serves a
like purpose very nearly as well.
152. — Floor-Beams. — The size of floor-beams can be as-
certained by the preceding rules for the stiffness of materials.
These rules give the required dimensions for the various
kinds of material in common use. The rules may be some-
what abridged for ordinary use, if some of the quantities
represented in the formula be made constant within certain
limits. For example, if the load per foot superficial upon
the floor be fixed, and the deflection, then these, together
with the constant 'represented by Fy may be reduced to one
126 CONSTRUCTION.
constant. For dwellings, the load per foot may be taken at
70 pounds, the weight proper to be allowed for a crowd
of people on their feet. (Transverse Strains, Art. 114.) To
this add 20 for the weight of the material of which the floor
is composed, and the sum, 90, is the value of f, or the weight
per foot superficial for dwellings. Then eft—-. U (Art. 130).
The rate of deflection allowable for this load may be fixed
at 0-03 inch per foot of the length, or d =. 0-03 /. Substitut-
ing these values in equation (35.), we obtain —
b d*= cfl* 9°c/3 = 1875 */3
i -6 Fx -03 / ~ ~ i -6 x -03 /^ ~ F
or —
. , ' i<r=2*ZLfr. (43.)
T X*7 C
Putting/ to represent — — , we have —
(44.)
T 8*7 C
Now, by reducing — •— -, for the six woods in common use,
the value of j for each is found as follows:
Georgia Pine ......................... / = 0-32
Locust ...... ......................... j = 0-37
White Oak.... ........................ j = 0-6
Spruce ............................... j — 0-54
White Pine ............ ' ............... j = 0-65
Hemlock .... .......................... j ^0-67
Equation (44.") is a rule for the floor-beams of dwellings ;
it may be used also to obtain the dimensions of beams for
stores for all ordinary business- for it will require from 3 to
5 times the weight used in this rule, or from 200 to 400
(average 300) pounds to increase the deflection to the limit
of elasticity in beams of the usual depths and lengths. For
light stores, therefore, loaded, say, to 150 pounds per foot,
the beams would be safe, but the deflection would be in-
CONSTANTS FOR USE IN THE RULES. 127
creased to 0-06 per foot. When so great a deflection as this,
would not be objectionable to the eye, then this rule (44.)
will serve for the beams of light stores. But for first-class
stores, taking the rate of deflection at -04 per foot, and *
fixing the weight per superficial foot at 275 pounds, includ-
ing the weight of the material of which the floor is con-
structed, and letting k represent the constant, then —
bd*=kcl\ (45.)
and for —
Georgia Pine k = 0-73
Locust k = 0-85
White Oak k = 1-38
Spruce k — 1-48
White Pine k = 1-23
Hemlock. . , k — 1-53
153. — Floor-Beam^ for Dwellings— To find the dimen-
sions of floor-beams for dwellings, when the rate of deflection"
is 0-03 inch per foot, or for ordinary stores when the load is
about 150 pounds per foot, and the deflection caused by this
weight is within the limits of the elasticity of the material,
we have the following rule :
Rule XXXVIIL— Multiply the cube of the length by
the distance apart between the beams (from centres), both in
feet, and multiply the product by the value of/ (Art. 152)
for the material of the beam, and the product will equal the
product of the breadth into the cube of the depth. Now,
to find the breadth, divide this product by the cube of the
depth in inches, and the quotient will be the breadth in
inches. But if the depth is sought, divide the said product
by the breadth in inches, and the cube root of the quotient
will be the depth in inches ; or if the breadth and depth are
to be in proportion as r is to unity, r representing any re-
quired decimal, then divide the aforesaid product by the
value of r, and extract the square root of the quotient, and
the square root of this square root will be the depth re-
quired in inches, and the depth multiplied by the value of i
will be the breadth in inches.
128 CONSTRUCTION.
Example. —In a dwelling Or ordinary stOre, what must be
the breadth of the beams, when placed 15 inches from
centres, to support a floor covering a span of 16 feet, the
depth being 1 1 inches, the beams of white oak ? By the
rule, 4096, the cube of the length, by i|, the distance from
centres, and by 0-6, the value of j for white oak, equals
3072. This divided by 1331, the cube of the depth, equals
2-31 inches, or 2T^ inches, the required breadth. But if, in-
stead of the breadth, the depth be required, the breadth
being fixed at 3 inches, then the product, 3072, as above, di-
vided by 3, the breadth, equals 1024 ; the cube root of this
is 10-08, or, say, 10 inches nearly. But if the breadth and
depth are to be in proportion, say, as 0-3 to i-o, then the
aforesaid product, 3072, divided by 0-3, the value of r,
equals 10240, the square root of which is 101-2, and the
square root of this is 10-06, the required depth. This
multiplied by 0-3, the value of r, equals 3-02, the re-
quired breadth ; the beam is therefore to be, say, 3 x 10
inches.
154. — Floor-Beams for First-Class Stores. — To find the
breadth and depth of the beams for a floor of a first-class
store sufficient to sustain 250 pounds per foot superficial
(exclusive of the weight of the material in the floor), with
a deflection of 0-04 inch per foot of the length, we have —
Rule XXXIX.— The same as XXXVIIL, with the ex-
ception that the value of k (Art. 152) is to be used instead
of the value of j.
Example. — The beams of the floor of a first-class store
are to be of Georgia pine, with a clear bearing between the
walls of 18 feet, and placed 14 inches from centres: what
must be the breadth when the depth is 1 1 inches ? By the
rule, 5832, the cube of the length, and i£, the distance from
centres, and 0-73, the value of k for Georgia pine, all multi-
plied together equal 4966-92 ; and this product divided by
1331, the cube of the depth, equals 3-732, the required
breadth, or 3| inches.
But if, instead of the breadth, the depth be required :
what must be the depth when the breadth is 3 inches ?
DISTANCE APART OF FLOOR-BEAMS. 129
The said product, 4966-92, divided by 3, the breadth, equals
1655-64, and the cube root of this, 11-83, or> sav> I2 inches,
is the depth required.
But if the breadth and depth are to be in a given pro-
portion, say 0-35 to i-o, the 4966-92 aforesaid divided by
0-35, the value of r, equals 14191, the square root of
which is 119-13, and the square root of this square root is
10-91, or, say, n inches, the required depth. And 10-91
multiplied by 0-35, the value of r, equals 3-82, the required
breadth, say 3^ inches.
I55« — Floor -Beams: Distance from Centres. — It is
sometimes desirable, when the breadth and depth of the
beams are fixed, or when the beams have been sawed and
are now ready for use, to know the distance from cen-
tres at which such beams should be placed in order that the
floor be sufficiently stiff. By a transposition of the factors
in equation (44.), we obtain —
bd*
In like manner, equation (45.) produces —
_bd*
(470
These, in words at length, are as follows :
Rule XL.— Multiply the cube of the depth by the breadth,
both in inches, and divide the product by the cube of the
length in feet multiplied by the value of /, for dwellings
and for ordinary stores, or by k, for first-class stores, and
the quotient will be the distance apart from centres in feet.
Example.— K span of 17 feet, in a dwelling, is to be cov-
ered by white-pine beams 3x12 inches: at what distance
apart from centres should they be placed? By the rule,
1728, the cube of the depth, multiplied by 3, the breadth,
equals 5184. The cube of 17 is 4913 ; this by 0-65, the value
of j for white pine, equals 3193-45- The aforesaid 5184
divided by this 3193-45 equals 1-6233 feet, or, say, 20 inches.
CONSTRUCTION.
156. — Framed Openings for Chimney* and Stairs. —
Where chimneys, flues, stairs, etc., occur to interrupt the
bearing, the beams are framed into a piece, b (Fig. 42), called
a header. The beams, a a, into which the header is framed
are called trimmers or carriage-beams. These framed beams
require to be made thicker than the common beams. The
header must be strong enough to sustain one half of the
weight that is sustained upon the &w7-beams, c c (the wall at
the opposite end or another header there sustaining the other
half), and the trimmers must each sustain one half of the
weight sustained by the header in addition to the weight it
supports as a common beam. It is usual in practice to make
these framed beams one inch thicker than the common beams
for dwellings, and two inches thicker for heavy stores. This
practice in ordinary cases answers very well, but in extreme
cases these dimensions are not proper. Rules applicable
generally must be deduced from the conditions of the case—
the load to be sustained and the strength of the material.
157. — Breadth of Headers. — The load sustained by. a
header is equally distributed, and is equal to the superficial
area of the floor supported by the header multiplied by the
load on every superficial foot of the floor. This is equal to
the length of the header multiplied by half the length of the
tail-beams, and by the load per superficial foot. Putting g
DIMENSIONS OF HEADERS. Ijl
for the length of the header, n for the length of the tail-
beams, and / for the load per superficial foot ; U, the uni-
formly distributed load carried by the header, will equal £
f n g. By substituting for £/, in equation (35.), this value of
it, we obtain —
The symbols g and / here both represent the same thing,
the length of the header ; combining these, and for # putting
its value gry we obtain—
.
3-2 Fr
To allow for the weakening of the header by the mor-
tices for the tail-beams (which should be cut as near the
middle of the depth of the header as practicable), the depth
should be taken at, say, one inch less than the actual depth.
With this modification, we obtain —
If /be taken at 90, and r at 0-03, we have, by reducing—
h _ 937- 5 "g* (4Q.)
-~'
which is a rule for the breadth of headers for dwellings and
for ordinary stores. This, in words, is as follows :
Rule XLL— Multiply 937-5 times the length of the tail-
beams by the cube of the'length of the header, both in feet.
The product divided by the cube of one less than the depth
multiplied by the value of F, Table III., will equal the
breadth of the header in inches for dwellings or ordinary
stores.
Example.— K header of white pine, for a dwelling, is 10
feet long, and sustains tail-beams 20 feet long ; its depth is
\2 inches: what must be its breadth? By the rule,
937. 5x20x10*= 18750000.' This divided by (12- i)3x 2900^
CONSTRUCTION.
3859900, equals 4-858, say 5 inches, the required breadth.
F or first-class stores,/ should be taken at 275, and r at 0-04.
With these values the constants in equation (48.) reduce to
2I48-4375, or, say, 2150. This gives—
a rule for the breadth of a header for first-class stores. It
is the same as that for dwellings, except that the constant
2150 is to be used in place of 937-5. Taking the same ex-
ample, and using the constant 2150 instead of 937-5, we
obtain 1 1 • 14 as the required breadth of the header for a first-
class store. Modifying the question by using Georgia pine
instead of white pine, we obtain 5 -476 as the required thick-
ness, say 5^ inches.
158. — Breadth of Carriage-Beams. — A carriage-beam or
trimmer, in addition to its load as a common beam, carries
one half of the load on the header, which, as has been
seen in the last article, is equal to one half of the superficial
area of the floor supported by the tail-beams multiplied by
the weight per superficial foot of the load upon the floor ;
therefore, when the length of the header in feet is repre-
sented by g, and the length of the tail-beams by n, w equals
- x - x /, equals £ f g n*
For a load not at middle, we have (25.) —
4 W ' amn
b d = —BJ—
*The load from the header, instead of being \fg n, is, more accurately,
i/«(g — c) : because the surface of floor carried by the header is only
that which occurs between the surfaces carried by the carriage-beams, each of
which carries so much of the floor as extends half way to the first tail-beam
from it, or the distance - ; therefore, the width of the surface carried equals
the length of the header less ( 2 x - = W, or g — c. When, however, it is con-
sidered that the carriage-beam is liable to receive some weight from a stairs or
other article in the well-hole, the small additional load above referred to is
not only not objectionable, but is really quite necessary to be included in the
calculation.
THICKNESS OF CARRIAGE-BEAMS. 133
This is a rule based upon resistance to rupture. By substi-
7? /
tuting for a, the factor of safety, -p-j-, its value in terms of
resistance to flexure {Transverse Strains, (154.)), we have —
h j* — 4 W B Im n __ 4 Wm n .
BIFdr Fdr
In this expression, W is a concentrated weight at the dis-
tances m and n from the two ends of the beam. Taking the
load upon a carriage-beam due to the load from the header,
as above found, and substituting it for W, we obtain —
,,* __ fgmn*
bd = --
This is the expression required for the concentrated load.
To this is to be added the uniformly distributed load upon
the carriage-beam ; this is given in equation (35.). Substi-
tuting for U of this equation its value, fc /, gives —
, ,,
-T67^" Fr
Combining these two equations, we have for the total load —
r r
If, in this equation, /be taken at 90, and r at 0-03, these
reduce to 3000 ; therefore, with this value of -, we have —
/ N
This rule for the breadth of carriage-beams with one
header, for dwellings and for ordinary stores, is put in words
as follows :
134 CONSTRUCTION.
Rule XLII. — Multiply the length of the framed opening
by its breadth, and by the square of the length of the tail-
beams ; to this product add f of the cube of the length into
the distance of the common beams from centres — all in feet ;
divide 3000 times the sum by the cube of the depth in inches
multiplied by the value of F for the material of the beam, in
Table III., and the quotient will be the breadth in inches.
Example. — In a tier of 3 x 10 inch beams, placed 14 inches
from centres, what should be the breadth of a Georgia-pine
carriage-beam 20 feet long, carrying a header 12 feet long,
.having tail-beams 1 5 feet long? Here the framed opening
is 5-x 1 2 feet. Therefore, according to the rule, 12 x 5 x 15' =
13500; to which add(|x2o3x jf =)5833i; the sum is 19333^,
and this by 3000= 58000000. The value of F for Georgia
pine, in Table III., is 5900; the cube of the depth is 1000;
the product of these two is 5900000; therefore, dividing
the above 58000000 by 5900000 gives a quotient of 9.83,
the required breadth in inches. If, in equation (51.), f be
taken at 275, and r at 0-04, then - becomes 6875, and the
equation becomes —
,
~~
Fd*
a rule for the breadth of carriage-beams for first-c lass stores ;
the same as that for dwellings, except that the constant is
6875 instead of 3000.
fl59. — Breadth of Carriage-Beams Carrying Two Sets
of Tail-Beams.— r-A rule for this is the same as that for a car-
riage-beam carrying one set of tail-beams, if to it there be
added the effect of the second set of tail-beams. Equation
(51.) with the additibn named becomes —
, ,
(54°
in which n is the length of one set of tail-beams, and s the
length of the other set ; and m + n = I.
CARRIAGE-BEAMS WITH TWO HEADERS. 135
If / be taken at 90, and r at 0-03, these two reduce to
3000, and we have —
_ 3000 [,£•;/ (;;?;; -Kr2)-f | r/3]
Pd*~
9
a rule for the breadth of a carriage-beam carrying two sets
of headers, for dwellings and for ordinary stores. It may
be stated in words as follows :
Rule XLIII. — Multiply the length of the longer set of
tail-beams by the difference between this length and the
length of the carriage-beam, and to the product add the
square of the length of the shorter set of tail-beams ; mul-
tiply the sum by the length of the longer set of tail-beams,
and by the length of the header ; to this product add f of
the product of the cube of the length of the carriage-
beam into the distance apart from centres of the common
beams ; multiply this sum by 3000 ; divide this product by
the product of the cube of the depth in inches into the value
of F ior the material of the carriage-beam, in Table III., and
the quotient will be the required breadth.
Example. — In a tier of 3 x 12 inch beams, placed 14 inches
from centres, what should be the breadth of a spruce car-
riage-beam 20 feet long in the clear of the bearings, carry-
ing two sets of tail-beams, one of them 9 feet long, the
other 5 feet ; the headers being 15 feet long ? The difference
between the longer set of tail-beams and the carriage-beam
is (20 — 9 =) 1 1 feet. Therefore, by the rule, 9 x 1 1 + 5* =
124; then (124x9x15=) 16740 + (f x 203 x |f —) 5833^- =
22573i; then 22573^x3000 = 67720000. Now the value of
F for spruce, Table III., is 3500; this by I23, the cube of
the depth, equals 6048000; by this dividing the aforesaid
67720000, we obtain a quotient of 11-197, the required
breadth of the carriage-beam. If, in equation (54.), / be
taken at 275, and > at 0-04, these reduce to 6875, and we
obtain —
a rule for the breadth of carriage -beams carrying two sets
CONSTRUCTION.
of tail-beams, in the floors of first-class stores. -This is like
the rule for dwellings, except that the constant is 6875 in-
stead of 3000.
160. _ Breadth o£ Carriage - Beam willi Well-Hole at
middle. — When the framed opening between the two sets of
tail-beams occurs at the middle, or when the lengths of the
two sets of tail-beams are equal, then equation (54.) reduces
to
and if /be taken at 90, and r at 0-03, these reduce to 3000,
and we have—
a rule for the breadth of a carriage-beam carrying two sets
of tail-beams of equal length, in the floor of a dwelling or of
an ordinary store ; and which in words is as follows:
Rule XLIV.— Multiply the length of the header by the
square of the length of the tail-beams, and to the product
add | of the product of the square of the length of the car-
riage-beam by the distance apart from centres of the com-
mon beams; multiply the sum by 3000 times the length of
the carriage-beam ; divide the product by the product of
the cube of the depth into the value of F for the material of
the carriage-beam, in Table III., and the quotient will be the
required breadth.
Example. — In a tier of 3x12 inch beams, placed 12 inches
from centres, what must be the thickness of a hemlock car-
riage-beam 20 feet long, carrying two sets of tail-beams,
each 8 feet long, with headers 10 feet long? By the rule,
10 x 8" + £ x i x 20* — 890 ; 890 x 3000 x 20 = 53400000. Now,
the value of F, in Table III., for hemlock is 2800 ; this by the
cube of the depth, 1728, equals 4838400; by this dividing
the former product, 53400000, and the quotient, 11-0367, is
the required breadth of the carriage-beam.
CROSS-BRIDGING.
If, in equation (57.), /be taken at 275, and rat 0.04, these
will reduce, to 6875, and we shall have —
6875
(59-)
a result the same as in equation (58.), except that the
constant is 6875 instead of 3000. Equation (59.) is a rule for
the breadth of carriage-beams carrying two sets of tail-beams
of equal length, in the floor of a first-class store. In words
at length, it is the same as Rule XLI V., except that the con-
stant 6875 is to be used in place of 3000.
161.— €ro§§-Bridgiiig, or Herring-Bone Bridging.— The
diagonal struts set between floor-beams, as in Fig. 43, are
known as cross- bridging, or herring-
bone bridging. By connecting the
beams thus at intervals, say, of from
5 to 8 feet, the stiffness of the floor
is greatly increased. The absolute
strength of a tier of beams to resist a
weight uniformly distributed over
the whole tier is augmented but lit-
tle by cross-bridging ; but the power
of any one beam in the tier to re-
sist a concentrated load upon it, as a heavy article of fur-
niture or an iron safe, is greatly increased by the cross-
bridging; for this device, by connecting the loaded beam
with the adjacent beams on each side, causes these beams to
assist in carrying the load. To secure the full benefit of the
diagonal struts, it is very important that the beams be well
secured from separating laterally, by having strips, such as
cross-furring, firmly nailed to the under edges of the beams.
The tie thus made, together with that of the floor-plank on
the top edges, will prevent the thrust of the struts from sep-
arating the beams.
162.— Bridging: Value to Resi§t Concentrated L,oad§.—
A rule for determining the additional load which any one
beam connected by bridging will be capable of sustaining,
by the assistance derived from the other beams, through the
FIG. 43.
138 CONSTRUCTION.
bridging, may be found in Chapter XVIII., Transverse Strains.
This rule may be stated thus : *
R = -(1 + 22 + 3V 42+ etc.) ; (60.)
in which R is the increased resistance, equal to the addi-
tional load which may be put upon the loaded beam ; c is the
distance from centres in feet at which the beams in the tier
are placed ; /is the load in pounds per superficial foot upon
the floor ; / is the length of the beams in feet ; and d is the
depth of the beams in inches. The squares within the
bracket are to be extended to as many places as there are
beams on each side which contribute assistance through the
bridging. The rule given in the work referred to, for ascer-
taining the number of spaces between the beams, is —
d f ,- \
« = 7r; (61.)
or, the depth of the beam in inches divided by the square of
the distance from centres, in leet, at which the beams are
placed will give the number of spaces between the beams
which contribute on each side in sustaining the concentrated
load. The nearest whole number, minus unity, will equal
the required number of beams.
The value of c for beams in floors of dwellings is given in
equation (46.), and lor those in first-class stores in equation
(47.). By a modification of equation (34.), putting c f I for
U, we have —
and-
c =
= (63.)
These equations give general rules for the value of c.
INCREASED LOAD BY CROSS-BRIDGING. 139
Now, the rule, in words at length, for the resistance offered
by the adjoining beams to a weight concentrated upon one
of the beams sustained by cross-bridging to the others, is —
Rule XLV. — Divide the depth of the beam in inches by
the square of the distance apart from centres in feet at
which the floor-beams are placed ; from the quotient deduct
unity, and call the whole number nearest to the remainder
the First Result. Take the sum of the squares of the con-
secutive numbers from unity to as many places as shall equal
the above first result ; multiply this sum by 5 times the
length in feet, by the load per foot superficial upon the floor,
and by the fifth power of the distance apart from centres in
feet at which the beams are placed ; divide the product by
4 times the square of the depth in inches, and the quotient
will be the weight in pounds required.
Example. — In a tier of 3 x 12 inch floor-beams 20 feet
long, placed in a dwelling 16 inches from centres and well
bridged, what load maybe uniformly distributed upon one of
the beams, additional to the load which that beam is capable
of sustaining safely when unassisted by bridging? Here,
according to the rule, 12 divided by (ij+ ij = ) i-J equals
6f ; 6J — i — 5|, the nearest whole number to which is 6, the
first result. The sum of the square of the first 6 numbers
equals (i + 22 + 3' + 4* + 5° + 62 =) I +4 + 9+ 16 + 25 + 36 = 91.
Therefore, 91 x 5 x 20 x 90 x (|-)r' = 345 1266.* The square of
the depth (12 x 12 = ) H4x4 = 576; by this dividing the
above 3451266, we have the quotient 5991-78, say 5992
pounds, the required weight. This is the additional load
which may be placed upon the beam. At 90 pounds per
superficial foot, the common load on each beam, we have
* The value of c, 16 inches, equals £ feet. The fifth power of this, or (£)6,
is obtained by involving both numerator and denominator to the fifth power,
and dividing the fifth power of the former by the fifth power of the latter ; for
(i)5 = i_. For the numerator we have 4x4x4x4x4=1024, and for the de-
35
nominator 3x3x3x3x3 = 243. The former divided by the latter gives as a
quotient 4-214, the value of (j)5. The process of involving a number to a high
power, or the reverse operation of extracting high roots, may be performed by
logarithms with great facility. (See Art. 427.)
I4O CONSTRUCTION.
90 x 20 x | = 2400 as the common load. To this add 5992,
the load sustained through the bridging by the other beams,
and the sum, 8392 pounds, will be the total load which may
be safely sustained, uniformly distributed, upon one beam —
nearly 3^ times the common load.
163.— Crirclers.— When the distance between the walls of
a building is greater than that which would be the limit for
the length of ordinary single beams, it becomes requisite to
introduce one or more additional supports. Where sup-
ports are needed for a floor and partitions are not desirable,
it is usual to use a large piece of timber called a girder, sus-
tained by posts set at intervals of from 8 to 1 5 feet ; or, when
posts are objectionable, a framed construction called a
framed girder (Art. 196) ; or an iron box called a tubular
iron girder (Art. 182). When a simple timber girder is used
it is advisable, if it be large, to divide it vertically from end
to end and reverse the two pieces, exposing the heart of the
timber to the aii» in order that it may dry quickly, and also
to detect decay at the heart. When the halves are bolted
together, thin slips of wood should be inserted between
them at the several points at which they are bolted, in order
to leave sufficient space for the air to circulate freely in the
space thus formed between them. This tends to prevent
decay, which will be found first at such parts as are not
exactly tight, nor yet far enough apart to permit the escape
of moisture. When girders are required for a long bear-
ing, it is usual to truss them ; that is, to insert between
the halves two pieces of oak which are inclined towards each
other, and which meet at the centre of the length of the
girder like the rafters of a roof-truss, though nearly if not
quite concealed within the girder. This and many similar
methods, though extensively practised, are generally worse
than useless ; since it has been ascertained that, in nearly all
such cases, the operation has positively weakened the girder.
A girder may be strengthened by mechanical contrivance,
when its depth is required to be greater than any one piece
of timber will allow. Fig. 44 shows a very simple yet invalu-
able method of doing this. The two pieces of which the gir-
CONSTRUCTION OF GIRDERS. 14!
der is composed are bolted or pinned together,, having keys
inserted between to prevent the pieces from sliding. The
keys should be of hard wood, well seasoned. The two
pieces should be about equal in depth, in order that the
joint between them may be in the neutral line. (See Arts.
120, 121.) The thickness of the keys should be about half
their breadth, and the amount of their united thickness
should be equal to a trifle over the depth and one third of
the depth of the girder. Instead of bolts orpins, iron hoops
are sometimes used ; and when they can be procured, they
are far preferable. In this case, the girder is diminished at
the ends, and the hoops driven from each end towards the
middle. A girder may be spliced if timber of a sufficient
length cannot be obtained ; though not at or near the mid-
FIG. 44-
die, if it can be avoided. (See Art. 87.) Girders should
rest from 9 to 12 inches on each wall, and a space should be
left for the air to circulate around the ends, that the damp-
ness may evaporate.
164.— Girders : IMmeniions.— The size of a girder, for
any special case, may be determined by equations (21.), (22.),
(25,), (27.), and (28.), to resist rupture ; and to resist deflection,
by equations (32.) and (35.). For girders in dwellings, equa-
tion (44.) may be used. In this case, the value of c is to be
taken equal to the width of floor supported by the girder,
which is equal to the sum of the distances half way to the
wall or next bearing on each side. When there is but one
142 CONSTRUCTION.
girder between the two walls, the value of c is equal to half
the distance between the walls. The rule for girders for
dwellings, in words, is —
Rule XLVL— Multiply the cube of the length of the gir-
der by the sum of the distances from the girder half way to
the next bearing on each side, and by the value of j for the
material of the girder, in Art. 152; the product -will equal
•the product of the breadth of the girder into the cube of the
depth. To obtain the breadth, divide this product by the
cube of the depth ; the quotient will be the breadth. To
obtain the depth, divide the said product by the breadth ;
the cube root of the quotient will be the depth. If the
breadth and depth are to be in a given proportion, say as
r : i-o, then divide the aforesaid quotient by the value of
r ; take the square root of the quotient; then the square root
of this square root will be the depth, and the depth multi-
plied by the value of r will be. the breadth.
Example. — In the floor of a dwelling, what should be the
size of a Georgia-pine girder 14 feet long between posts,
placed at 10 feet from one wall and 20 feet from the other?
The value of c here is V~ + %° — ¥ = J5- The value of j
for Georgia pine (Art. 152) is 0-32. By the rule, 14' x 15 x
0-32 = 13171-2. Now, to find the breadth when the depth
is 12 inches ; 13171 -2 divided by the cube of 12, or by 1728,
gives a quotient of 7-622, or 7$, the required breadth.
Again,, to find the depth, when the breadth is 8 inches :
13171-2 divided by 8 gives 1646-4, the cube root of which is
1 1 -808, or, say, uj inches, the required depth. But if
neither breadth nor depth have been previously determined,
except as to their proportion, say as 0-7 to i-o, then
13171-2 divided by 0-7 gives 18816, of which the square
root is 137-171, and of this the square root is 11-712, or, say,
I if inches, the required depth. For the breadth, we have
11-712 by 0-7 equals 8-198, or, say, 8J, the required
breadth. Thus the girder is required to be 7$- x 12, 8 x iij,
or 8J xi if inches. This example is one in a dwelling or
ordinary store ; (or first-class stores the rule for girders is the
same as the last, except that the value of k is to be taken
instead ofy, in Art. 152.
FIRE-PROOF TIMBER FLOORS. 143
165. — Solid Timber Floors. — Floors constructed with
rolled-iron beams and brick arches are proof against fire
only to a limited degree ; for experience has shown that the
heat, in an extensive conflagration, is sufficiently intense to
deprive the iron of its rigidity, and consequently of its
strength. Singular as it may seem, it is nevertheless true
that wood, under certain circumstances, has a greater fire-
resisting quality than iron. Floors of timber constructed,
as is usual, with the beams set apart, have but little power
to resist fire, but if the spaces between the beams be filled
up solid with other beams, which thus close the openings
against the passage of the flames, and the under surface be
coated with plastering mortar containing a large portion of
plaster of Paris, and finished smooth, then this wooden
floor will resist the action of fire longer than a floor of iron
beams and brick arches. The wooden beams should be se-
cured to each other by dowels or spikes.
166. — Solid Timber Floors for Dwellings and Assem-
bly-Rooms. — From Transverse Strains, Art. 702, we have —
which may be modified so as to take this form :
which is a rule for the depth or thickness of solid timber
floors for dwellings, assembly-rooms, or office buildings, and
in which y and h are constants depending upon the mate-
rial ; thus, for —
Georgia Pine ................ y — 4» and h =
Spruce ...................... ^ = 2i, " // = 0-365
White Pine .................. y = 2j, " // = 0-389
Hemlock ..................... y = 2, ' 7*=o-39
The rule may be stated in words thus :
Rule XLVIL— Multiply the length by the value of
144 CONSTRUCTION.
and by the value of /i, as above given ; to the product add
82 ; multiply the sum by the cube of the length ; divide this
product by 0-576 times the value of F, in Table III. ; then
the cube root of the quotient will be the required depth in
inches.
Example. — What depth is required for a solid Georgia-
pine floor to cover a span of 20 feet ? For Georgia pine
F= 5900 ; y, as above given, equals 4, and h equals 0-314 ;
therefore, by the rule —
d - _ = 6.318;
0-576x5900 3398'4
or, the depth required is, say, 6-32 or 6^ inches.
167. — Solid Timber Floors for First-Clas§ Si ores. — The
equation given for first-class stores, in Transverse Strains,
Art. 702, is —
» -(263
which may be changed to this form :
in which y is as before, and k for —
Georgia Pine equals 0-4
Spruce equals o • 472
White Pine equals 0-502
Hemlock equals o • 506
This rule may be put in words the same as Rule XLVIL,
except as to the constants, which require that 263 be used in
place of 82, that k be used in place of /*, and that 0-768 be
used in place of 0-576. Table XXI. of Transverse Strains
contains the results of computation showing the depths of
solid timber floors for dwellings and assembly-rooms and
for first-class stores, in floors of spans varying from 8 to 30
feet, and for the four kinds of timber before named.
IRON FLOOR-BEAMS. 145
I6ff. — Rolled - Iron Beam*. — The dimensions of iron
beams, whether Avrought or cast, are to be ascertained by
the rules already given, when the beams are of rectangular
form in their cross-section; these rules are applicable alike
to wood and iron (Art. 93), and may be used for any mate-
rial, provided the constant appropriate to the given mate-
rial be used. But when the form of cross-
section is such as that which is usual for
rolled-iron beams (Fig. 45), the rules
need modifying. Without' attempting
to explain these modifications (referring
for this to Transverse Strains, Art. 457
and following article), it may be re-
marked that the elements of resistance
to flexure in a beam constitute what is
termed the Moment of Inertia. This, in
. FIG. 45.
a beam of rectangular cross-section, is
equal to -J^ of the breadth into the cube of the depth ; or —
I=^bd\ (66.)
This would be appropriate to rolled-iron beams if the
hollow on each side were filled with metal, so as to complete
the form of cross-section into a rectangle. The proper ex-
pression for them may be obtained by taking first the
moment for the beam as if it were a solid rectangle, and
from this deducting the moment for the part which on each
side is wanting, or for the rectangles of the hollows. In
accordance with this view of the case, we have —
b,d^; (67.)
in which b is the breadth of the beam or width of the
flanges ; b, is the breadth of the two hollows, or is equal to#
less the thickness of the web or stem ; d is the depth includ-
ing top and bottom flanges ; and dt is the depth in the clear
between the top and bottom flanges.
Now, if equation (32.) be divided by 12, we shall have —
146 CONSTRUCTION.
and since -^ b d* represents the moment of inertia, we have-
(68.)
12 Fd
This gives the value of / for a beam of any form in cross-
section loaded at the middle. By this equation the values
of / have been computed for rolled -iron beams of many
sizes, and the results recorded in Table XVII., Transverse
Strains. A few of these are included in Table IV., as follows :
TABLE IV. — ROLLED- IRON BEAMS.
NAME.
Depth.
Weight
per yard.
/ =
NAME.
Depth.
Weight
per yard.
/ =
Trenton
qo
7-8d
Buffalo
QO
109- 1 17
Paterson . .
Phoenix
5
t
3°
36
12-082
I4'3I7
Phoenix
Buffalo
T?*
150
QO
190-63
151 -436
Trenton
•6
-1O
°v 761
Buffalo
ioi
IO^
17^ -6-1^
Phoenix . .
Trenton . .
Buffalo
7
•7
8
55
60
65
42-43
46-012
64- 526
Trenton
iBuffalo
Paterson. . . .
1
loj
12$
I2±
135
125
125
241-478
286-OI9
292-05
Paterson. . .
Phoenix ... .
Phoenix . . .
8
9
9
So
?o
84
84-735
92-207
107-793
Paterson. ...
iBuffalo
Trenton
12*
12*
!5fV
170
1 80
150
398-93^
418-945
528-223
169. — Rolled -Iron Beams: Diiuciitioii* ; Weight at
middle. — If, in equation (68.), there be substituted for F its
value for wrought iron, as in Table TIL, we shall have —
or —
" 1 2 x 62000 <y '
Wl*
744000 rf
(69.)
This is a rule by which to ascertain the size of a rolled-iron
beam to sustain a given weight at middle with a given de-
flection, and, in words at length, is as follows :
Rule XLVII1. — Multiply the weight in pounds by the
cube of the length in feet ; divide the product by 744000
times the deflection in inches, and the quotient will be the
DEFLECTION OF IRON BEAMS. 147
moment of inertia of the required beam, and may be found,
or the next nearest number, in Table IV. in column headed
7. Opposite to the number thus found, to the left, will be
found the name, -depth, and weight per yard of the required
beam.
Example. — Which of the beams of Table IV. would be
proper to carry 10,000 pounds at the middle with a deflection
of one inch, the length between bearings being 20 feet ?
Here we have, substituting for the symbols their values —
W Y3 _ioooox2o3 _ 80000000
~ 744000 tf ~~~ 744000 x i ~~ 744000 3 ' '
or, the moment of inertia of the required beam is 107-527, the
nearest to which, in the table, is 107-793, pertaining to the
Phoenix 9-inch, 84-pound beam. This, then, is the required
beam.
170. — Rolled-Iron Beams: Deflection when Weight is
at Middle. — By a transposition of symbols in equation (69.),
we have —
W lz
~-~ (70.)
or a rule for the deflection of rolled-iron beams when the
weight is at the middle. This, in words, is—
Rule XLIX.— Multiply the weight in pounds by the
cube of the length in feet ; divide the product by 744000
times the value of / for the given beam, and the quotient
will be the required deflection in inches.
Example.-JN\Mi\. will be the deflection of a Phoenix 9-
inch, 70-pound beam 20 feet long, loaded at the middle with
7500 pounds ? The value of / for this beam, in Table IV.,
is 92-207; therefore, substituting for the symbols their val-
ues, and proceeding by the rule, we have—
__ ___. =
" 744000 / " " 744000 x 92 - 207
or, the deflection will be, say, f of an inch.
148 CONSTRUCTION.
171. — Rolled -Iron Beams : Weight when at middle.—
A transposition of factors in equation (70.) gives —
IV— 744QQQ / <?
/3 (71.)
This is a rule for the weight at middle, and, in words, is —
Rule L. — Multiply 744000 times the value of 7 by the
deflection in inches ; divide the product by the cube of the
length, and the quotient will be the required weight in
pounds.
Example. — What weight at the middle of a Buffalo g-inch,
go-pound beam will deflect it one inch, the length between
bearings being 20 feet ? The value of 7 for this beam, in
Table IV., 18109-117; therefore—
744000 /<? 744000x109-117x1
^ 203 : IO147 -WSJ
or, the required weight is, say, 10,148 pounds.
172. — Rolled -Iron Beam§ : Weight at any Point. — The
equation for a load at any point is (Transverse Strains, Art.
485)-
186000 IS
—r^r,r - (72.)
in which m and n represent the two parts m feet into which
the point where the load rests divides the length. This, in
words, is as follows :
Rule LI. — Multiply 186000 times the value of 7 by the
deflection in inches ; divide the product by the product of
the length into the rectangle formed by the two parts into
which the point where the load rests divides the length ;
the quotient will be the required weight in pounds.
Example. — What weight is required, located at 10 feet
from one end, to deflect i£ inches a Paterson 12^-inch, 125-
pound beam 25 feet long between bearings ? The value of
7 for this beam, in Table IV., is 292-05 ; m — 10, and n = I
- m = 25 — 10 ^ 15 ; therefore —
VARYING WEIGHTS ON IRON BEAMS. 149
,,, I86OOO Id I86OOO X 2Q2-O5 X I -s
- -
or, the required weight is, say, 21,730 pounds.
(73. — Rolled-Iron Beams: Dimension* • Weight at any
Point.— By transposition of factors in equation (72.), we ob-
tain —
Wlmn
'" 186000 tf* (73-)
This may be expressed in words as follows :
Rule LI I. — Multiply the weight by the length, and by the
rectangle of the two parts into which the point where the
weight rests divides the length ; divide the product by 186000
times the deflection, and the quotient will be the value of 7,
which (or its next nearest number) may be found in Table
IV., opposite to which will be found the required beam.
Example. — What beam 10 feet long will be required to
carry 5000 pounds at 3 feet from one end with a deflection
of 0-4 inch? Here we have m equal 3, and n equal 7;
therefore —
Wlmn 5000x10x3x7
- . U ' - - J A . J I •?
1 86000 tf " 1 86000 x o • 4
The value of /is 14-113, the nearest number to which in
the table, is 14-317, the moment of inertia of the Phoenix 5-
inch, 36-pound beam ; this, therefore, is the .beam required.
174. — Rolled-Iron Beams: Dimensions; Weight Uniform-
ly Distributed.— Since f U = W(Art. 138), equation (69.) may
be modified by the substitution of this value of W, when we
obtain —
which reduces to —
7 =
744000 $'
/ = — p (74.)
1 190400 6
150 CONSTRUCTION.
a rule for the dimensions of a beam for a uniformly distrib-
uted load, which, in words, is as follows :
Rule LIII. — Multiply the uniformly distributed load by
the cube of the length ; divide the product by 1 190400 times
the deflection, and the quotient will be the value of /, corre-
sponding to which, or to its next nearest number will be
found in Table IV/the required beam.
Example. — What beam 10 feet long is required to sus-
tain an equally distributed load of 14,000 pounds with a de-
flection of half an inch ? For this we have—
14000 x io8
1190400x0-5
This is the moment of inertia of the required beam ; nearly.
the same as 23-761, in Table IV., the value of / fora Tren-
ton 6-inch, Ao-pound beam, which will serve as the re-
quired beam.
175. — Rolled-Iron Beam* : Deflection ; Weight Uniformly
Distributed. — A transposition of the factors in equation (74.)
gives —
*= UIS
11904007' (75-)
a rule for the deflection of a uniformly loaded beam, and
which may be put in these words, namely :
Rule LIV. — Multiply the uniformly distributed load by
the cube of the length ; divide the product by 1190400 times
the value of /, Table IV., and the quotient will be the re-
quired deflection.
Example. — To what depth will 14,000 pounds, uniformly
distributed, deflect a Buffalo lo^-inch, 9O-pound beam 20
feet long? The value of /for this beam, as per the table, is
151- 436 ; therefore —
14000 X 20 3
—?•=• 0-6213 ;
1 190400 x 151 -436
or, the required deflection is, say, •£• of an inch.
IRON FLOOR-BEAMS FOR DWELLINGS. 151
! 7 6.— Rolled -Iron Beam§: Weight when Uniformly
Distributed. — Equation (75.), by a transposition of factors,
gives—
*9 (76.)
a rule for the weight uniformly distributed, and which may
be worded thus :
Rule LV. — Multiply 1190400 times the value of /, Table
IV., by the deflection ; divide the product by the cube of
the length, and the quotient will be the required weight.
Example. — What weight uniformly distributed upon a
Buffalo loj-inch, 105 -pound beam 25 feet long between
bearings will deflect it f of an inch ?
The value of / for this beam, as per Table IV., is 175-645 ;
therefore —
1190400 17
or, the required weight is, say, 10,036 pounds.
(77. — Rolled-Iron Beam§: Floors of Dwellings or As-
semfoly- Rooms.— From Transverse Strains, Art. 500, we
have —
a rule for the distance from centres of rolled-iron beams in
floors of dwellings, assembly-rooms, or offices, where the
spaces between the beams are filled in with brick arches and
concrete. In the equation, c is the distance apart from cen-
tres in feet, and y is the weight per yard of the beam. This,
in words, is thus expressed :
Rule LVL— Divide 255 times the value of / by the cube
of the length ; from the quotient deduct one 42oth part of
the weight of the beam per yard, and the remainder will
the required distance apart from centres.
Example.— What should be the distance apart from cen-
1 5 2 CONSTRUCTION.
tres of Buffalo I2j-inch, 125 -pound beams 25 feet lo.ng
between bearings, in the floor of an assembly-room ? For
these beams, in Table IV., / equals 286-019, and y= 125;
therefore—
_ 255 x 286-oi9_ 125 m
~^J™ ~ 420 '
- -298 = 4-37;
15625 420
or, the required distance from centres is, say, 4 feet 4^
inches.
178. — Rolled-Iron Beams : Floor§ of First-Class Stores.
— From Transverse Strains, Art. 504, we have —
a rule for the distance from centres of rolled-iron beams in
the floor of a first-class store ; the spaces between the beams
being filled with brick arches and concrete. This rule may
be put in words as follows:
Rule LVIL— Divide 148-8 times the value of 7 by the
cube of the length ; from the quotient deduct one 96oth
part of the weight of the beam per yard, and the remainder
will be the distance apart of the beams from centres in
feet.
Example. — What should be the distance apart from cen-
tres of Buffalo I2j-inch, i8o-pound beams 20 feet long
between bearings, in the floor of a first-class store? For
these beams the value of /, Table IV., is 418-945, and the
value of y is 180; therefore—
c = I48'8 x 418-945 _ 180 6o .
203 960
or, the required distance from centres is, say, 7 feet /|
inches.
TIE-RODS FOR IRON BEAMS. 153
179. — Floor- A roll es : Oeneral Considerations. — In fill-
ing the spaces between the iron beams of a floor, the arches
should be constructed with hard whole brick of good shape,
laid upon the supporting centre in contact with each other,
and the joints thoroughly filled with cement grout, and
keyed with slate. Made in this manner, the arches need not
be over four inches thick at the crown for spans extending
to 7 or 8 feet, and 8 inches thick at the springing, where
they should be started upon a proper skew-back. The rise
of the arch should not be less than i£ inches for each foot
of the span.
180. — Floor - Arches ; Tie -Rods: Dwellings. — From
Transverse Strains ', Art. 507, we have —
d = V 0-0198 c s, (79.)
which is a rule for the diameter in inches of a tie-rod for
an arch in the floor of a bank, office building, or assembly-
room ; in which d is the diameter in inches of the rod, s is
the span of the arch, and c is the distance apart between the
rods (s and c both in feet). This rule requires that the arch
rise i^ inches per foot of the span, and that the brick-work
and the superimposed load each weigh 70 pounds, or to-
gether 140 pounds. This rule, in words, is -as follows:
Rule LVIII. — Multiply the span of the arch by the dis-
tance apart at which the rods are placed, and by the decimal
0-0198 ; the square root of the product will be the diameter
of the required rod.
Example. — What should be the diameter of the wrought-
iron ties of brick arches of 5 feet span, in a bank or hall of
assembly, where the ties are 8 feet apart ? For this we
have —
d — 1/0-0198 x^8 x 5 = 1/-792 = 0-89;
or, the diameter of the required rods should be, say, -J of an
inch.
181. — Floor-Arches ; Tie-Rods: First-Class Stores. — From
the same source as in last article, we have —
d = V o- 04527 c~s, (80.)
154
CONSTRUCTION.
which is a rule for the size of tie-rods for the brick arches
of the floors of first-class stores, where the arches have a
rise of i£ inches for each foot of the span, and where the
weight of the brick arch and concrete is not over 70 pounds
per superficial foot of the floor, and the loading does not
exceed 250 pounds per superficial foot. As the rule is the
same as the one in the preceding article, except the deci-
mal, a recital of the rule, in words, is not here needed. To
obtain the required diameter, proceed as directed in Rule
LVIII., using the decimal 0-04527 instead of the one there
given.
TUBULAR IRON GIRDERS.
182. — Tubular Iron Girders: De§cripfion. — The use of
wooden beams for floors is limited to spans of about 25 feet.
When greater spans than this are to be covered, some expe-
dient must be resorted to by which
intermediate bearings for the floor-
beams may be provided. Wooden
girders may be used, but these
need to be supported by posts at
intervals of from 10 to 15 feet,
unless the girders are trussed, or
made up of top and bottom chords,
struts, and ties. And even this
is objectionable, owing to the
height such a piece of framing
requires, and which encumbers
the otherwise free space of the
hall. A substitute for the framed
girder has been found in the
tubular iron girder, as in Fig. 46, made of rolled plate
iron and angle irons, riveted. They require to be stiffened
by an occasional upright T iron along eachside, and a
cross-head at least at each bearing.
183. — Tubular Iron Girders: Area of Flangc§ ; Load
at Middle. — In wrought-iron tubular girders it is usual to
make the top and bottom flanges of equal thickness. From
Transverse Strains, Art. 551, we have —
FIG. 46.
TUBULAR IRON GIRDERS. 1 55
a rule for the area of the bottom flange ; in which a' equals
the area of the flange in inches, W the weight in pounds at
the middle, / the length and d the depth of the girder, both
in feet, and k the saTe load in pounds per inch with which
the metal may be loaded, and which is usually taken at
9000. The rule may be stated thus :
Rule LIX. — Multiply the weight by the length ; divide
the product by 4 times the depth into the value of k, and
the quotient will be the required area of the bottom flange.
Example. — In a girder 40 feet long and 3 feet high, to
carry 75,000 pounds at the middle, what area of metal is
required in the bottom flange, putting k at 9000? For this
we have, by the rule —
. W I 75000 x 40
f,' — — j__j_ ~ , — ^7 •77'*
4 d k 4 x 3 x 9000
or, the area required is 27$ inches. This is the amount of
uncut metal. An allowance is required for that which will
be cut by rivet-holes. This is usually an addition of one
sixth.
184. Tubular Iron Girders : Area of Flanges; Load at
any Point.— The equation suitable for this (Transverse
Strains, Art. 553) is —
(82-}
in which m and n are the distances respectively from the lo-
cation of the load to the two ends of the girder. The other
symbols are the same as in the last article. This rule may
be thus stated :
Rule LX.— Multiply the weight by the values of m and
of n ; divide the product by the product of the depth into
the length and into the value of k, and the quotient will be
the required area of the bottom flange.
Example.— In a girder 50 feet long between bearings and
156 CONSTRUCTION.
3J- feet high, what area of metal is required in the bottom
flange to sustain 50,000 pounds at 20 feet from one end, when
k equals 9000 * By the rule, we have —
rr m n 50000 x 20 x 30
or, each flange requires 19 inches of solid metal uncut for
rivets.
185. — Tubular Iron Girder§ : Area of Flange§ ; Load
Uniformly Distributed. — The equation appropriate here is
(Transverse Strains, Art, 555) —
This is a rule by which to obtain the area of cross-sec-
tion of the bottom flange at any point in the length of the
girder, the load uniformly distributed ; m and n being the
respective distances from the point measured to the two
ends of the girder, and U representing the uniformly dis-
tributed load in pounds. This, in words, is described as
follows :
Rule LXI. — Divide the weight by the product of twice
the depth into the length and into the value of k ; then the
quotient multiplied by the values of m and of n will be the
required area of the bottom flange at the point measured,
the distance of which from the ends equals m and n.
Examplc.~\n a girder 50 feet long and 3^ feet high, to
carry a uniformly distributed load of 120,000 pounds, what
area of cross-section is required in the bottom flange, at the
middle and at intervals of 5 feet thence, to each support ; k
being taken at 9000? Here we have, first—
. m n 1 20000 m n
a = U — -jT-r = — - = 0-038005 m n.
2dkl 2x3^x9000x50
Now, when m = n = 25, we have the middle point ; then —
a' = 0-038095 m n = 0-038095 x 25 x 25 = 23-81 ;
/Til w T i
SHEARING STRAIN.
j& "^
\vv
SRJ^r
or, the area of the bottom flange at mid-length is 23
inches.
When ;;/ = 20, then n = 30, and —
a' = 0-038095 X2ox 30 = 22-86;
or, the required area, at 5 feet either way from the middle,
is 22-J inches.
When m = 15, then n = 35, and —
a = 0-038095 x 15 x 35 = 20-0 ;
or, at 10 feet either way from the middle, the required area
is 20 inches.
When m= 10, then n = 40, and —
a' — 0-038095 x 10x40 = 15-24 ;
or, at 15 feet either way from the middle, the required area
is 15^ inches.
When m = 5, then n — 45, and —
a' = 0-038095 x 5x45 =8-57;
or, at 20 feet each side of the middle, the required area is 8f
inches.
The area of cross-section found in every case is that of
the uncut fibres ; to this is to be added as much as will be
cut by the rivets. This is usually about one sixth of the area
given by the rule. The top flange is to be made equal in
area to the bottom flange. The flanges are unvarying in
width from end to end, the variation of area being obtained
by varying the thickness of the flanges, and this being at-
tained by building the flange in lamina, or plates ; but these
should not be less than a quarter of an inch thick. There
should be added to the length of the girder, in the clear,
about one tenth of its length for supports on the walls : thus,
a girder 30 feet long requires 3 feet added for supports, or
1 8 inches on each wall.
186.— Tubular Iron Girdcr§: Shearing Strain.— The top
and bottom flanges are provided of sufficient size to resist
158 CONSTRUCTION.
the transverse strain ; the two upright plates, technically
termed the web, need, therefore, to be thick enough to resist
only the shearing strain. This, upon a beam uniformly
loaded, is at the middle theoretically nothing, but from
thence it increases regularly towards each support, where it
equals half the whole weight. For example, the girder of
Art. 185, 50 feet long between supports, carries 120,000
pounds uniformly distributed over its length. In this case
the shearing strain at the wall at each end is the half of
120,000 pounds, or 60,000 pounds; at 5 feet from the wall it
is -fg or -J less, or 48,000 pounds ; at 10 feet from the wall it
is f less, or 36,000 pounds ; at 15 feet it is 24,000 ; at 20 feet
it is 12,000; and at 25 feet or the middle, it is nothing.
187.— Tubular Iron Girder§: Thickness of Wefo.— The
equation appropriate for this is —
in which t is the thickness of the web (equal to the sum of
the thicknesses of the two side plates), d is the height of the
plate (/ and d both in inches), G is the shearing strain, and k'
is the effective resistance of wrought iron to shearing per
inch of cross-section. This may be put in words as follows :
Rule LXII. — Divide the shearing strain by the product of
the depth in inches into the value of /£', and the quotient
will be the thickness of the web, or of the two side plates
taken together.
Example. — What is the required thickness of web in a
girder 50 feet between bearings, side plates 38 inches high
between top and bottom flanges, and to carry 120,000 pounds,
uniformly distributed ? Here, putting the shearing resistance
of the plates at 7000 pounds per inch, we have —
dk' ' 38x7000 266000*
The shearing strain at the supports, as in last article, is
60000 ; therefore, we have for this point—
LIGHT IRON GIRDERS. 159
60000
1 "266000" ^ °'225'
When G = 48000, then—
48000
: "^660^0"- °'I8;
and when G = 36000, then—
36000
/ — -?— — 0-135.
266000
Those nearer the middle of the girder are still less than
these ; and these are all below the practicable thickness,
which is half an inch for the two plates. The plates ought
not in practice ever to be made less than a quarter of an
inch thick.
188. — Tubular Iron Girder§, for Floors of
A§§cmbly-Room§, and Office Buildings.— When the floors of
these buildings are constructed with rolled-iron beams and
brick arches, then the following (Art. 568, Transverse Strains]
is the appropriate equation for the area of cross-section of
the bottom flange of the girder :
in which a' is in inches, and c, c , d, /, m, and n are in feet.
Also, a' is the area required ; y is the weight per yard of the
rolled-iron beam of the floor ; c, their distances from centres ;
c', the distance from centres at which the tubular girders are
placed, or the breadth of floor carried by one girder; d, the
depth of the girder; k, the effective resistance of the metal
per inch in the flanges of the girder ; and ;// and n are the
distances respectively from the two ends of the girder to the
point at which the area of cross-section of the bottom flange
is required. The rule may be thus described :
Rule LXIIL— Divide the weight per yard of the rolled-
iron beams by 3 times their distance from centres; to the
quotient add 140 and reserve the sum ; deduct the length
in feet from 700, and with the remainder as a divisor divide
700 ; multiply the quotient by the above reserved sum, and
l6o CONSTRUCTION.
by the value of c' ; divide the product by the product of
twice the depth into the value of k, and the quotient multi-
plied by the values of m and of n will be the required area
of cross-section of the bottom flange at the point in the
length distant from the two ends equal to m and n respec-
tively.
Example. — In a floor of 9-inch, /o-pound beams, 4 feet
from centres, what ought to be the area of the bottom flange
of a tubular girder 40 feet long between bearings, 2 feet 8
inches deep, and placed 17 feet from the walls or from other
girders ; the area of the flange to be ascertained at every 5
feet of the length : the value of k to be put at 9000 ? Here
y •=. 70, c = 4, cf = 17, 7=40, and d= 2f. Therefore, by
the rule —
/ 70 \ 700 17
a = I 14.0+ ) x o x m n :
V 3 x 47 700 - 40 2 x 2f x 9000
a' — 145 • 8^- x I -0606 x 0-0003 54rf x mn\
a' = 0-05478 m n.
The values of m and n are —
At the middle m = 20 ; n = 20
5 feet from middle m— 15; n = 25
10 " " " m •=• 10 ; n = 30
15 . " " " m= 5 ; n = 35
These give —
At the middle a' — 0-05478 x 20 x 20 = 21 -91
" 5 feet from middle a' = 0-05478 x 15 x 25 = 20-54
"10 " " " a' = 0-05478 x 10 x 30 — 16-43
"15 " " " *' = 0-05478 x 5x35= 9-59
These are the areas of uncut fibres at the points named, in
the lower flange ; the upper flange requires the same sizes.
(89. — Tubular Iron Girders, for Floors of First-Class
Stores.— The equation proper for this is (Transverse Strains,
Art. 570)—
CAST IRON COMPARED WITH WROUGHT. l6l
a rule the same in form as that of the previous article ; hence
it needs no particular exemplification.
Rule LXIII. of last article may be used for this case,
simply by using the constant 320 in place of that of 140.
CAST-IRON GIRDERS.
190. — Cast-iron Girders: Inferior. — Rolled- iron beams
have been so extensively introduced within a few years as
to have superseded almost entirely the formerly much used
cast-iron beam or girder. The tensile strength of cast iron
is far inferior to that of wrought iron. This inferiority and
the contingencies to which the metal is subject in casting
render it very untrustworthy ; it should not be used where
rolled-iron beams can be procured. A very substantial gir-
der to carry a brick wall is made by placing two or more
rolled-iron beams side by side, and securing them together
by bolts at mid-height of the web ; placing thimbles or sep-
arators at each bolt. As there may be cases, however, in
which cast-iron girders will be used, a few rules for them
will here be given.
1 9 1. _Cast-Iron Girder: Load at Middle. — The form of
cross-section given to this girder usually is as shown in
Fig. 47.
In the cross-section, the bottom flange
is made to contain in area four times as
much as the top flange. The strength
will be in proportion to the area of the
bottom flange, and to the height or
depth of the girder at middle. Hence,
to obtain the greater strength from a
given amount of material, it is requisite
to make the upright part, or the web,
rather thin ; yet, in order to prevent
injurious strains in the casting while it
is cooling, the parts should be nearly
equal in thickness. The thickness of the three parts— web,
1 62 CONSTRUCTION.
top flange, and bottom flange — may be made in proportion
as 5, 6, and 8.
For a weight at middle, the form of the web should be
that of a triangle ; the top flange forming two straight lines
declining from the centre each way to the bottom flange at
the ends, like the rafters of a roof to its tie-beam. From
Transverse Strains, Art. 583, we have —
which is a rule for the area in inches of the bottom flange,
for a load at middle ; the area of the top flange is to be equal
to one fourth of that of the bottom flange. To secure this,
make the width of the top flange equal to one third of the
width of the bottom flange ; the thickness of the former,
as before directed, being made equal to f or f of the lat-
ter. The weight W is in pounds ; the length / is in feet ;
and the depth d is in inches. The factor of safety a should
be taken at not less than 3 ; better at 4 or 5.
The equation in words may be as follows :
Rule LXIV.— Multiply the weight by the length, and by
the factor of safety ; divide the product by 4850 times the
depth at middle, and the quotient will be the area in inches
of the bottom flange ; divide this area by the width of the
bottom flange, and the quotient will be its thickness. Of the
top flange make its width equal one third that of the bot-
tom flange, and its thickness equal to three quarters that of
the latter. Make the thickness of the web equal to f- that
of the bottom flange.
Example. — What should be the dimensions of the cross-
section of a cast-iron girder 20 feet long between bearings,
and 24 inches high at middle, where 30,000 pounds is to be
carried ; the factor of safety being put at 5 ?
Here we have W — 30000 ; a — 5 ; / — 20 ; and d =
24 ; therefore, by the rule—
30000 x 5 x 20
THE BOWSTRING GIRDER.
I63
This is the area of the bottom flange. If the width of this
flange be 12 inches, then 25-773 divided by 12 gives 2 -15, or
2-J- full, as the thickness. One third of 12 equals 4, equals
the width of the top flange ; and j- of 2 • 1 5 equals i • 6 1 , or i f —
its thickness. The thickness of the web equals -| x 2-15 =
i -34 or i-J inches.
192 — Cast-iron Girder: Load Uniformly Distributed.
The equation suitable to this is —
Ual
9700 (f
(88.)
a rule of like form with that of the last article ; therefore,
Rule LXIV. may be used for this case, simply by substitut-
ing 9700 for 4850.
193.— Cast -Iron Bowstring Girder.— An arched girder,
such as that in Fig. 48, is technically termed a " bowstring
girder." The curved part is a cast-iron beam of T form in
section, and the horizon-
tal line is a wrought-iron
tie-rod attached to the
ends of the arch. This
girder has but little to
commend it, and is by no
means worthy the confi-
dence placed in it by FlG- 48'
builders, with many of whom it is quite popular. The brick
arch usually turned over it is adequate to sustain the entire
compressive force induced from the load (the brick wall
built above it), and it thereby supersedes the necessity for the
iron arch, which is a useless expense. The tie-rod is the
only useful part of the bowstring girder, but it is usually
made too small, and not infrequently is seriously injured by
the needless strain to which it is subjected when it is
" shrunk in" to the sockets in the ends of the arch. The bow-
string girder, therefore, should never be used.
(94.— Substitute for tlie Bowstring Girder. — As the cast-
iron arch of a bowstring girder serves only to resist com-
164
CONSTRUCTION.
pression, its place can as well be filled by an arch of brick,
footed on a pair of cast-iron skew-backs ; and these held
in position by a pair of wrought-iron tie-rods, as shown in
Fig. 49. This system of construction is preferable to the
bowstring girder, in that the
tie-rods are not liable to injury
by " shrinking in," and the
cost is less. From Transverse
Strains, Art. 596, we have —
D =
Ul
9425 d
(89.)
FIG. 49.
an equation in which D is the
diameter in inches of each of
the two tie-rods of the brick
arch ; U is the load in pounds
uniformly distributed over the arch ; / is the span of the
arch in feet ; and d, in inches, is its versed sine, or its height
measured from the centre of the tie-rod to the centre of the
thickness or height of the arch at middle.
This equation may be put in words as follows :
Rule LXV. — Multiply the weight by the length ; divide
the product by 9425 times the depth, and the square root of
the quotient will be the diameter of each rod.
Example. — What should be the diameter of each of the
pair of tie-rods required to sustain a brick arch 20 feet span
from centres, with a versed sine or height at middle of 30
inches, to carry a brick wall 12 inches thick and 30 feet high,
weighing TOO pounds per cubic foot? The load upon this
arch will be for so much of the wall as will occur over the
opening, which will be about one foot less than the span of
the arch, or 20 — i = 19 feet. Therefore, the load will
equal 19 x 3ox i x 100 = 57,000. pounds ; and hence, U —
57000, / = 20, d — 30, and, by the rule —
57000 x 20
__. -- = 1/4-0318 = 2. 008;
9425 x 30
or, the diameter of each rod is required to be 2 inches.
STRAINS REPRESENTED GRAPHICALLY. 165
FRAMED GIRDERS.
195 — Graphic Representation of Strain§ In the first
part of this section, commencing at Art. 71, the metfiod was
developed of ascertaining the strains in the various parts of
a frame by the parallelogram or triangle of forces. The
method, so far as there explained, is adequate to solve sim-
ple cases ; but when more than three pieces of a frame con-
verge in one point, the task bv that method becomes difficult.
This difficulty, however, disappears when recourse is had to
the method known as that of 4< Reciprocal Figures, Frames,
and Diagrams of Forces," proposed by Professor I. Clerk
Maxwell in 1867. This is an extension of the method by
the triangle of forces, and may be illustrated as follows :
FIG. 50. FIG. 51.
Let the lines in Fig. 50 represent, in direction and
amount, four converging forces in equilibrium in any frame,
as, for example, the truss of a roof; let the lines in Fig. 51
be drawn parallel to those in Fig. 50, in the manner fol-
lowing, namely : Let the line A B be drawn parallel with the
line of Fig. 50 which is between the corresponding letters
A and B, and let it be of corresponding length ; from B draw
the line B C parallel with the line of Fig. 50 which is be-
tween the letters B and C, and of corresponding length ;
then from C draw CD, and from A draw A D, respectively
parallel with the lines of Fig. 50 designated by the corre-
sponding letters, and extend them till they intersect at D.
The lengths of these two lines, the last two drawn, are de-
termined by the point D where they intersect; their lengths,
therefore, need not be previously known,. The lengths of
the lines in Fig. 51 are respectively in proportion to the
l66 CONSTRUCTION.
several strains in Fig. 50, provided these strains are in
equilibrium. Fig. 5 1 is termed a closed polygon of forces.
A system of such polygons, one for each point, in the frame
where forces converge, so constructed that no line repre-
senting a force shall be repeated, is termed a diagram of
forces. This diagram of forces is a reciprocal of the frame
from which it is drawn, its lines and angles being the same.
The facility of tracing the forces in the diagram of forces
depends materially upon the system of lettering here shown,
and which was proposed by Mr. Bow, in his excellent work
on the Economics of Construction. In this system each
line of the frame is designated by the two letters which it
separates ; thus the line between A and B is called line A B ;
that between C and D is called line CD ; and so of others ;
and in the diagram the corresponding lines are called by the
same letters, but here the letters designating the line are, as
usual, at the ends of the line. Any point in a frame where
forces converge is designated by the several letters which
cluster around it; as, for example, in Fig. 50, the point of
convergence there shown is designated as point A B C D.
This invaluable method of denning graphically the
strains in the various pieces composing a frame, such as a
girder or roof-truss, is remarkably simple, and is of general
application. Its utility will now be exemplified in its appli-
cation to framed girders, and afterwards to roof-trusses.
196.— Framed 4*irder§. — Girders of solid timber are use-
ful for the support of floors only where posts are admissible
as supports, at intervals of from 8 to 15 feet. For unob-
structed long spans it becomes requisite to construct a frame
to serve as a girder (Arts. 163, 182). A frame of this kind
requires two horizontal pieces, a top and a bottom chord,
and a system of struts and suspension-pieces by which
the top and bottom chords are held in position, and the
strains from the load are transmitted to the bearings at the
ends of the girders. Various methods of arranging these
struts and ties have been proposed. One of the most simple
and effective is shown in Fig. 52, forming a series of isos-
celes triangles. The proportion between the length and
height of a girder is important as an element of economy
RULES FOR FRAMED, GIRDERS. 167
both of space and cost. When circumstances do not control
in limiting- the height, it may be determined by this equation
from Transverse Strains, Art. 624 —
, (I75+/)/
2400 ; (90.)
in which d is the depth or height between the axes of the
top and bottom chords, and / is the length between the cen-
tres of bearings at the supports (d and / both in feet). This
equation in words is as follows :
Rule LXVI. — To the length add 175 ; multiply the sum
by the length ; divide the product by 2400, and the quotient
will be the required height between the axes of the top and
bottom chords.
Example. — What should be the depth of a girder which
is 40 feet long between the centres of action at the supports?
For this the rule gives—
(175+40^40 s
2400
or, the proper depth for economy of material is 3 feet and
7 inches.
The number of bays, panels, or triangles into which the
bottom chord may be divided is a matter of some considera-
tion. Usually girders from—
20 to 59 feet long should have 5 bays.
59 « 85 " " " " 6 "
85 " 107 " " " " 7 "
107 " 127 " " " " 8 "
127 " 146 " " " 9
(97.— Framed Girder and Diagram of Force§. — Let Fig.
52 represent a framed girder of six bays of, say, n feet
each, or of a total length of 66 feet.
The lines shown are the axial lines, or the imaginary lines
passing through the axes of the several pieces composing the
frame. The. six arrows indicate the six pressures into which
the equally distributed load is supposed to be divided. Each
of these is at the apex of a triangle, the base of which 1
along the lower chord.
1 68
CONSTRUCTION.
The spaces between the arrows are lettered ; so, also, the
space between the last arrow at either end and the point of
support has a letter, and so has each triangle, and there is
one for the space beneath the lower chord. These letters
are to be used in describing the diagram of forces, as was
explained in Art. 195. The diagram of forces (Fig. 53) for
this girder-frame is drawn as follows, namely : Upon a verti-
FIG. 52.
cal line A ^Vmark the points A, O, P, Q, R, S, and N, at equal
distances, to represent the six equal vertical pressures indi-
cated by the arrows in Fig. 52. The equal distances A O,
OP, etc., may be made of any convenient size; but it will
HGrF
FIG. 53.
serve to facilitate the measurement of the forces in the dia-
gram if they are made by a scale of equal parts, and the
number of parts given to each division be made equal to the
number of tons of 2000 pounds each which is contained in
the pressure indicated by each arrow. On this vertical line
the distance A 0 represents the load at the apex of the tri-
angle B, or the point A OCB (Art. 195); the distance OP
GRAPHICAL DIAGRAMS OF FORCES. 169
represents the weight at the second arrow, or at the point
O PR D C, and so of the rest. If the weights upon the
points in the upper chord had been unequal, then the divi-
sion of the vertical line A N would have had to be corre-
spondingly unequal, each division being laid off by the scale,
to accord with the weight represented by each. The line
of loads, A N, being adjusted, the other lines are drawn from
it (Art. 195), so as to make a closed polygon for the forces
converging at each point of the frame, Fig. 52— commenc-
ing with the point A B T, Fig. 52, where there are three
forces, namely, the force acting through the inclined strut
A B, the horizontal force in B 7", and the vertical reaction
A T at the point of support. This last is equal to half the
entire load, or equal to the pressure indicated by the three
arrows, A O, O P, and P Q, and is represented in Fig. 53 by
A Q or A T. From the point Q draw a horizontal line Q B ;
this is parallel with the force B T of Fig. 52, in the lower
chord. From the point A draw A B parallel with the strut
A B of Fig. 52. This line intersects the line B T in B and
closes the polygon A B TA ; the point B defines the length
of the lines A B and B 7", and these lines measured by the
scale by which the line of loads was constructed give the
required pressures in the corresponding lines, A B and B T,
of Fig. 52.
Taking next, the point ABCO, where four forces meet,
of which we already have two, namely, the force in the
strut A B and the load A O — from the point O draw the hori-
zontal line O C ; this is parallel to the horizontal force O C
of Fig. 52. Now from B draw B C parallel with the suspen-
sion-piece B C of Fig. 52. This line intersects O C in C, and
the point C limits the lines O C and B C and closes the poly-
gon A B C O A, the four sides of which are respectively in
proportion to the four forces converging at the point A B CO
of Fig. 52, and when measured by the scale by which the
line of loads was constructed give the required strains re-
spectively in each. Taking next the point B C D T, where
four forces converge, of which we already have two, B C
and B T— from B extend the horizontal line TB to D\ from
C draw CD parallel with CD of Fig. 52, and extend it to in-
tersect TD in D, and thus close the polygon T B C D T.
I/O CONSTRUCTION.
The lines in a part of this polygon coincide — those from
B to T\ this is because the two strains B T and D T, Fig. 52,
lie in the same horizontal line. Again, taking the point
OC D EP, where five forces meet, three of which, O P, O C,
and CD, we already have — draw from D the line D E parallel
with D E of Fig. 52, and from P the line PE horizontally or
parallel with P E of Fig. 52. These two lines intersect at E
and close the polygon PO C D E P, the sides of which meas-
ure the forces converging in the point PO CD E, Fig: 52.
Next in order is the point D E F T, Fig. 52, where four forces
meet, two of which, T D and D E, are known. From E draw
EF parallel with EF'm Fig. 52; and from 71, TF parallel
with TFin Fig. 52 ; these two lines meet in F and close the
polygon TD E F T, the sides of which measure the required
strains in the lines converging at the point DEFT, Fig. 52.
Taking next the point PEFG Q9Fig. 52, where five forces
meet, of which we already have three, QP, P£,and E F—
Irom F draw a line parallel with F G of Fig. 52, and from Q
a line parallel with Q G of Fig. 52. These two intersect at G
and complete the polygon QPEFG Q, the lines of which
measure the forces converging at PEFG Q in Fig. 52.
In this last polygon, a peculiarity seems to indicate an
error: the line FG has no length ; it begins and ends at the
same point ; or, rather, the polygon is complete without it.
This is easily understood when it is considered that the two
lines FG and G H do not contribute any strength towards
sustaining the loads P Q and QR, and in so far as these
weights are concerned they might be dispensed with, and
the space occupied by the three triangles F , G, and H left
free, and be designated by only one letter instead of three.
Thus it appears that there are only four instead of five forces
at the point PEFG Q, and that the four are represented by
the lines of the polygon QPEFQ.
The peculiarity above explained arises from considering
loads only on the top chord : the analysis of the case is cor-
rect as worked from the premises given ; but in practice
there is always more or less load on the bottom chord at the
middle, which should be considered. This will be included
in a case proposed in the next article. One half of the dia-
LOAD ON BOTH CHORDS.
I/I
gram of forces is now complete. The other half being ex-
actly the same, except that it is in reversed order, need not
here be drawn.
198. — Framed Girders: Load on Both Chords Let
Fig. 54 represent the axial lines of a girder carrying an
FIG. 54.
equally distributed load on each chord, represented by the
arrows and balls shown in the figure. Let each bay measure
IO feet, or the length ot the girder be 50 feet, and its height
/\7
XXX
M
FIG. 55-
be 4J- feet. The diagram of forces (Fig. 55) for this girder
is obtained thus :
The plan of the girder, Fig. 54> requires to be lettered
as shown ; having one letter within each panel and outside
the frame, and one between every two weights or strains.
Then, in Fig. 55, mark the vertical line K V at L, M, N, V,
1 72 CONSTRUCTION.
and Pt dividing it by scale into equal parts, corresponding
with the weights on the top chord represented by the ar-
rows. For example, if the load at each arrow equals 6J
tons, make K L, L M, M N, etc., each equal to 6^ parts of the
scale. Then K P will equal the total load on the top flange.
Make the distance P V equal to the sum of the loads on the
bottom chord. Then K V equals the total load on the gir-
der. Bisect K Fin U\ then K U or U Fequals half the total
load ; consequently, equals the reaction of the bearing at K
or P of Fig. 54.
Now, to obtain the polygon of forces converging at
K A U, Fig. 54, we have one of these forces, K U, or the re-
action of the bearing at KA U, equal to K U, Fig. 55. From
f/draw UA parallel with U A of Fig. 54, and from 1C draw
KA parallel with the strut KA, Fig. 54, and intersecting the
line UA at A, a point which marks the limit of K A and UA,
and closes the polygon K A UK, the sides of which are in
proportion respectively to the three strains which converge
at the point A UK, Fig. 54. For example, since the line K U
by scale measures the vertical reaction, K U, of the bearing
at A UK, Fig. 54, therefore the line K A of the diagram of
forces by the same scale measures the strain in the strut KA,
Fig. 54, and the line A U of the diagram by the same scale
measures the strain in the bottom chord at A U, Fig. 54. For
the strains converging at K A B L, Fig. 54, of which two,
KA and K L, are already known, we draw from A the line
A B parallel with the line A B, Fig. 54, and from L draw L B
parallel with L B, Fig. 54, meeting A B at B, a point which
limits the two lines and closes the polygon K A B L K, the
lines of which are in proportion respectively to the strains
converging at the point KA B L, Fig. 54, as before explained.
Of the five strains converging at U A B C T, we already have
three — T U, UA,and AB* to obtain the other two, make
UQ equal to PV, equal to the total load upon the lower
flange ; divide U Q into four equal parts, QR, RS, S T, and
T U, corresponding with the four weights on the lower
chord, and represented by the four balls, Fig. 54. Now,
from T, the point marking the first of these divisions, draw
TC parallel with T C, Fig. 54, and from B draw B C paral-
VARIOUS STRAINS IN FRAMED GIRDERS. 1/3
lei with the strut EC, Fig. 54, meeting TC in C, a point
which limits the lines B C and TC and closes the polygon
T U A B C T, the sides of which are in proportion respectively
to the strains converging in the point T U A B C T, Fig. 54.
Of the five forces converging at MLB CD, we already have
three— ML, LB, and B C\ to obtain the other two, from M
draw M D parallel with M D, Fig. 54, and from C draw CD
parallel with CD, Fig. 54, meeting MD at D, a point limit-
ing the lines M D and CD and closing the polygon
MLBCDM,i\\Q sides of which are in proportion to the
strains converging at the point MLB CD, Fig. 54. Of the
five forces converging at the point 5 TCD E, three — S T,
T C, and CD — are known; to obtain the other two, from 5
draw SE parallel with SE, Fig. 54, and from D draw D E,
parallel with the strut D E, Fig. 54, meeting the line SE in
Et a point limiting the two lines S E and D E and closing the
polygon 5 TCD E S, the sides of which are in proportion to
the strains converging at 5 TCDE, Fig. 54. One half of the
strains in Fig. 54 are now shown in its diagram of forces, Fig.
55 ; and since the two halves of the girder are symmetrical,
the forces in one half corresponding to those in the other,
hence the lines of the diagram for one half of the forces
may be used for the corresponding forces of the other half.
199. — Framed Girders : Dimensions of Parl§. — The
parts of a framed girder are the two horizontal chords (top
and bottom) and the diagonals— the struts and ties. The top
chord is in a state of compression, while the bottom chord
experiences a tensile strain. Those of the diagonal pieces
which have a direction from the top to the bottom chord,
and from the middle towards one of the bearings of the
girder, as KA, B C, or D E, Fig. 54, are struts, and are sub-
jected to compression. The diagonal pieces which have a
direction from the bottom to the top chord, and from the
middle towards one of the supports, as A B or CD, Fig. 54.
are ties, and are subjected to extension, (Art. 83). The
amount of strain in each piece in a framed girder having
been ascertained in a diagram of forces, as shown in Arts.
197 and 198, the dimensions of each piece may be obtained
174 CONST RUCTION.
by rules already given. The dimensions of the pieces in a
state of compression are to be ascertained by the rules for
posts in Arts. 107 to 114, and those in a state of tension by
A:mts. 117 to 119 (see Arts. 226 to 229). Care is required, in
obtaining the size of the lower chord, to allow for the joints
which necessarily occur in long ties, for the reason that tim-
ber is not readily obtained sufficiently long without splicing.
Usually, in cases where the length of the girder is too great
to obtain a bottom chord in one piece, the chord is made up
of vertical lamina, and in as long lengths as practicable, and
secured with bolts. A chord thus made will usually require
about twice the material ; or, its sectional area of cross-sec-
tion will require to be twice the size of a chord which is in
one whole piece ; and in this chord it is usual to put the fac-
tor of safety at from 8 to 10.
The diagonal ties are usually made of wrought iron, and
it is well to secure the struts, especially the end ones, with
iron stirrups and bolts. And, to prevent the evil effects of
shrinkage, it is well to provide iron bearings extending
through the depth of each chord, so shaped that the struts
and rods may have their bearings upon it, instead of upon
the wood.
PARTITIONS.
200 — Partitions — Such partitions as are required for
the divisions in ordinary houses are usually formed by tim-
ber of small size, termed studs or joists. These are placed
upright at 12 or 16 inches from centres, and Avell nailed.
Upon these studs lath are nailed, and these are covered
with plastering. The strength of the plastering depends in
a great measure upon the clinch iormed by the mortar which
has been pressed through between the lath. That this
clinch may be interfered with in the least possible degree, it
is proper that the edges of the partition-joists which are
presented to receive the lath should be as narrow as prac-
ticable ; those which are necessarily large should be reduced
by chamfering the corners. The derangements in floors,
plastering, and doors which too frequently disfigure the
interior of pretentious houses with gaping cracks in the
FRAMED PARTITIONS.
175
plastering and in the door-casings are due in nearly all cases
to defective partitions, and to the shrinkage of floor-timbers.
A plastered partition is too heavy to be trusted upon an ordi-
nary tier of beams, unless so braced as to prevent its weight
from pressing upon the beams. This precaution becomes es-
pecially important when, in addition to its own weight, the
partition serves as a girder to carry the weight of the floor-
beams next above it. In order to reduce to the smallest
practicable degree the derangements named, it is important
that the studs in a partition should be trussed or braced so
as to throw the weight upon firmly sustained points in the
construction beneath, and that the timber in both partitions
and floors should be well seasoned and carefully framed.
To avoid the settlement due to the shrinkage of a tier of
beams, it is important, in a partition standing over one in the
story below or over a girder, that the studs pass between
the beams to the plate of the lower partition, or to the
girder ; and, to be able to do this, it is also important to ar-
range the partitions of the several stories vertically over
each other. All principal partitions should be of brick,
especially such as are required to assist in sustaining the
floors of the building.
FIG. 56.
201.— Examples of Partition &.—Fig. 56 represents a par-
tition having a door in the middle. Its construction is simple
but effective. Fig. 57 shows the manner of constructing a
I76
CONSTRUCTION.
partition having- doors near the ends. The truss is formed
above the door-heads, and the lower parts are suspended
from it. The posts a and b are halved, and nailed to the
tie c d and the sill ef. The braces in a trussed partition
r
a b
FIG 57-
should be placed so as to form, as near as possible, an angle
of 40 degrees with the horizon. The braces in a partition
should be so placed as to discharge the weight upon the
FIG. 58.
points of support. All oblique pieces that fail to do this
should be omitted.
When the principal timbers of a partition require to be
large for the purpose of greater strength, it is a good plan
WEIGHT UPON PARTITIONS. 177
to omit the upright filling-in pieces, and in their stead to
place a few horizontal pieces, as in Fig. 58, in order that upon
these and the principal timbers upright battens may be
nailed at the proper distances for lathing. A partition thus
constructed requires a little more space than others ; but it
has the advantage of insuring greater stability to the plas-
tering, and also of preventing to a good degree the conver-
sation of one room from being overheard in the adjoining
one. Ordinary partitions are constructed with 3x4, 3x5,
or 4x6 inch joists, for the principal pieces, and with 2x4,
2x5, or 2x6 filling-in studs, well strutted at intervals of
about 5 feet. When a partition is required to support, in
addition to its own weight, that of a floor or some other
burden resting upon it, the dimensions of the timbers should
be ascertained, by applying the principles which regulate
the laws of pressure and those of the resistance of timber,
as explained in the first part of this section, and in Arts. 196
to 199 for framed girders. The following data may assist in
calculating the amount of pressure upon partitions :
White-pine timber weighs from 22 to 32 pounds per cubic
loot, varying in accordance with the amount of seasoning it
has had. Assuming it to weigh 30 pounds, the weight of
the beams and floor-plank in every superficial foot of the
flooring will be —
6 pounds when the beams are 3 x 8 inches, and placed 20 inches from centres
7\ " " " 3 x 10 " " 18
9 " " 3 x 12 " " 16
II " « « 3XI2 " 12
I3 " " " 4X12 " " 12
I3 « « « 4x14 " 14
In addition to the beams and plank, there is generally the
plastering, of the ceiling of the apartments beneath, and some-
times the deafening. Plastering may be assumed to weigh 9
pounds per superficial foot, and deafening 1 1 pounds.
Hemlock weighs about the same as white pine. A parti-
tion of 3x4 joists of hemlock, set 12 inches from centres,
therefore, will weigh about 2j pounds per foot superficial
and when plastered on both sides, 2o£ pounds.
CONSTRUCTION.
ROOFS.
202. — Roof*. — In ancient Norman and Gothic buildings,
the walls and buttresses were erected so massive and firm
that it was customary to construct their roofs without a tie-
beam, the walls being abundantly capable of resisting the
lateral pressure exerted by the rafters. But in modern
buildings, usually the walls are so slightly built as to be in-
capable of resisting much if any oblique pressure ; hence
the necessity of care in constructing the roof so as to avoid
oblique and lateral strains. The roof so constructed, instead
of tending to separate the walls, will bind and steady them.
FIG. 59.
FIG. 60.
FIG. 61.
FIG. 62.
FIG. 63.
FIG. 64,
FIG. 65.
FIG. 66.
FIG. 67.
203. — Comparison of Roof-Tru§ses. — Designs for roof-
trusses, illustrating various principles of roof construction,
are herewith presented.
The designs at Figs. 59 to 63 are distinguished from those
at Figs. 64 to 67 by having a horizontal tie-beam. In the
latter group, and in all designs similarly destitute of the
horizontal tie at the foot of the rafters,- the strains are much
greater than in those having the tie, unless the truss be pro-
VARIOUS FORMS OF ROOF-TRUSSES. 179
tected by exterior resistance, such as may be afforded by
competent buttresses.
To the uninitiated it may appear preferable, in Fig. 64,
to extend the inclined ties to the rafters, as shown by the
dotted lines. But this would not be beneficial ; on the con-
trary, it would be injurious. The point of the rafter where
the tie would be attached is near the middle of its length,
and consequently is a point the least capable of resisting
transverse strains. The weight of the roofing itself tends to
bend the rafter ; and the inclined tie, were it attached to the
rafter, would, by its tension, have a tendency to increase this
bending. As a necessary consequence, the feet of the rafters
would separate, and the ridge descend.
In Fig. 65 the inclined ties are extended to the rafters ;
but here the horizontal strut or straining beam, located at
the points of contact between the ties and rafters, counteracts
the bending tendency of the rafters and renders these points
stable. In this design, therefore, and only in such designs, it
is permissible to extend the ties through to the rafters.
Even here it is not advisable to do so, because of the in-
creased strain produced. (See Figs. 77 and 79.) The design
in Fig. 64, 66, or 67 is to be preferred to that in Fig. 65.
204.— Force Diagram : L,oacl upon Each Support.— By a
comparison of the force diagrams hereinafter given, of each
of the foregoing designs, we* may see that the strains in the
trusses without horizontal tie-beams at the feet of the rafters
are greatly in excess of those having the tie. In constructing
these diagrams, the first step is to ascertain the reaction of,
or load carried by, each of the supports at the ends of the
truss. In symmetrically loaded trusses, the weight upon
each support is always just one half of the whole load.
205.— Force Diagram for Trus§ in Fig. 59. — To obtain
the force diagram appropriate to the design in Fig. 59, first
letter the figure as directed in Art. 195, and as in Fig. 68.
Then draw a vertical line, EF (Fig. 69), equal to the weight W
at the apex of roof ; or (which is the same thing in effect)
equal to the sum of the two loads of the roof, one extending
on each side of W half-way to the foot of the rafter. Di-
1 8o
CONSTRUCTION.
vide E F into two equal parts at G. Make G C and G D eacTi
equal to one half of the weight N. Now, since £ G is equal
to one half of the upper load, and G D to one half of the low-
er load, therefore their sum, E G + G D — ED, is equal to
one half of the total load, or to the reaction of each support,
E or F. From D draw DA parallel with DA of Fig. 68, and
from E draw EA parallel with EA of Fig. 68. The three
lines of the triangle A E D represent the strains, respectively,
in the three lines converging at the point A D E of Fig. 68.
Draw the other lines of the diagram parallel with the lines of
c-
FIG. 69.
Fig. 68, and as directed in Arts. 195 and 197. The various
lines of Fig. 69 will represent the forces in the corresponding
lines of Fig. 68 ; bearing in mind (Art. 195.) that while a line
in the force diagram is designated in the usual manner by the
letters at the two ends of it, a line of the frame diagram is
designated by the two letters between which it passes. Thus,
the horizontal lines A D, the vertical lines A B, and the in-
clined lines A E have these letters at their ends in Fig. 69,
while they pass between these letters in Fig. 68.
206.— Force Diagram for Truss in Fig. 60. — For this truss
we have, in Fig. 70, a like design, repeated and lettered as
required. We here have one load on the tie-beam, and three
loads above the truss: one on each rafter and one at the
ridge. In the force diagram, Fig. 71, make G H, H J, and
J K, by any convenient scale, equal respectively to the
weights GH, HJ, and J K oi Fig. 70. Divide G K into two
equal parts at L. Make LE and ZFeach equal to one half
the weight E F (Fig. 70). Then G Fis equal to one half the
FORCE DIAGRAMS OF TRUSSES. jgl
total load, or to the load upon the support G (Art. 205).
Complete the diagram by drawing its several lines parallel
with the lines of Fig. 70, as indicated by the letters (see Art.
205), commencing with G F, the load on the support G (Fig.
70). Draw from F and G the two lines FA and GA paral-
lel with these lines in Fig. 70. Their point of intersection
defines the point A. From this the several points B, C, and
D are developed, and the figure completed. Then the lines
m Fig. 71 will represent the forces in the corresponding lines
of Fig. 70, as indicated by the lettering. (See Art. 195.)
K A
FIG. 70.
FIG. 71.
207. — Force Diagram for Tru§§ in Fig 61. — For this truss
we have, in Fig. 72, a similar design, properly prepared by
weights and lettering ; and in Fig. 73 the force diagram ap-
propriate to it.
In the construction of this diagram, proceed as directed
in the previous example, by first constructing N S, the ver-
tical line of weights ; in which line NOyOP,PQ,QR, and R S
are made respectively equal to the several weights above
the truss in Fig. 72. Then divide NS into two equal parts at T.
Make 7^ and TL each equal to the half of the weight K L.
Make J K and L M equal to the weights J K and L M of Fig.
72. Now, since M " N is equal to one half of the weights above
the truss plus one half of the weights below the truss, or half of
the whole weight, it is therefore the weight upon the support
N (Fig. 72), and represents the reaction of that support. A
horizontal line drawn from M will meet the inclined line
drawn from N, parallel with the rafter A N (Fig. 72), in the
1-82
CONSTRUCTION.
point^, and the three sides of the triangle A MN, Fig. 73, will
give the strains in the three corresponding lines meeting at
the point A MN, Fig. 72. The sides of the triangle HJ 5, Fig.
FIG. 72.
73, give likewise the strains in the three corresponding lines
meeting at the point H J 5, Fig. 72. Continuing the con-
struction, draw all the other lines of the force diagram parallel
FIG. 73.
with the corresponding lines of Fig. 72, and as directed in
Art. 195. The completed diagram will measure the strains
in all the lines of Fig. 72.
FORCE DIAGRAMS CONTINUED. 183
208,— Force Diagram for Truss in Fig. 63.— The roof
truss indicated at Fig. 63 is repeated in Fig. 74, with the ad
FIG. 74.
FIG. 75-
dition of the lettering required for the construction of
force diagram, Fig. 75.
1 84 CONSTRUCTION. .
In this case there are seven weights, or loads, above the
truss, and three below. Divide the vertical line O V at W
into two equal parts, and place the lower loads in two equal
parts on each side of W. Owing to the middle one of these
loads not being on the tie-beam with the other two, but on
the upper tie-beam, the line G H, its representative in the
force diagram, has to be removed to the vertical BJ, and
the letter M is duplicated. The line NO equals half the
whole weight of the truss, or 3^ of the upper loads, plus one
of the lower loads, plus half of the load at the upper tie-
beam. It is, therefore, the true reaction of the support NO,
and A N is the horizontal strain in the beam there. It will
be observed also that while H ' M and G M (Fig. 75), which
are equal lines, show the strain in the lower tie-beam at the
middle of the truss, the lines C H and FG, also equal 'but
considerably shorter lines, show the strains in the upper
tie-beam. Ordinarily, in a truss of this design, the strain in
the upper beam would be equal to that in the lower one,
which becomes true when the rafters and braces above the
upper beam are omitted. In the present case, the thrusts of
the upper rafters produce tension in the upper beam equal
to C M or F M of Fig. 75, and thus, by counteracting the
compression in the beam, reduce it to C H or FG of the
force diagram, as shown.
209. — Force Diagram for Truss in Fig. 64. — The force
diagram for the roof-truss at Fig. 64 is given in Fig". 77,
while Fig. 78 is the truss reproduced, with the lettering
requisite for the construction of Fig. 77.
The vertical E F (Fig. 77) represents the load at the
ridge. Divide this equally at W, and place half the lower
weight each side of W, so that CD equals the lower
weight. Then ED is equal to half the whole load, and
equal to the reaction of the support E (Fig. 76). The lines
in the triangle A D E give the strains in the corresponding
lines converging at the point A D E of Fig. 76. The other
lines, according to the lettering, give the strains in the cor-
responding lines of the truss. (See Art. 195.)
FORCE DIAGRAMS CONTINUED.
i85
210. — Force Diagram for Tru§s in Fig. 65. — This truss is
reproduced in Fig. 78, with the letters proper for use in the
force diagram, Fig. 79.
<
FIG. 79.
Here the vertical G K, containing the three upper loads
GH, HJ, and J K, is divided equally at W, and the lower
1 86 CONSTRUCTION.
load E F is placed half on each side of W, and extends from E
to F. Then FG represents one half of the whole load of the
truss, and therefore the reaction of the support G (Fig. 78).
Drawing the several lines of Fig. 79 parallel with the corre-
sponding lines of Fig. 78, the force diagram is complete, and
the strains in the several lines of 78 are measured by the cor-
responding lines of 79. (See Art. 195.)
A comparison of the force diagram of the truss in Fig. 76
with that of the truss in Fig. 78 shows much greater strains
in the latter, and we thus see that Fig. 76 or 64 is the more
economical form.
FIG. 80.
211. — Force Diagram for Truss in Fig. 66. — This truss is
reproduced and prepared by proper lettering in Fig. 80, and
its force diagram is given in Fig. 81.
Here the vertical J M contains the three upper loads
JK, KL, and LM. Divide 7 J/ into two equal parts at
G, and make FG and G H respectively equal to the two
loads FG and GH of Fig.Zo. Then HJ represents one
half of the whole weight of the truss, and therefore the reac-
tion of the support J. From H and J draw lines par-
allel with A H and A J of Fig. 80, and the sides of the tri-
angle A H J will give the strains in the three lines concen-
trating in the point A H J (Fig. 80). The other lines of Fig,
EFFECT OF ELEVATING THE TIE-BEAM. 187
8 1 are all drawn parallel with their corresponding lines in
Fig. 80, as indicated by the lettering. (See Art. 195.)
FIG. 81.
212 Roof-Truss: Effect of Elevating tbe Tie-Beam.—
From Arts. 670, 671, Transverse Strains, it appears that the
FIG. 83
effect of substituting inclined ties for the horizontal tie at
feet of rafters is—
1 88 CONSTRUCTION.
in which P represents half the weight of the whole truss and
the load upon it ; a + & = height of the truss at middle above
a horizontal line drawn at the feet of the rafters ; a equals
the height from this line to the point where the two inclined
ties meet ; b, the height thence to the top of the truss ; and
F, the additional vertical strain at the middle of the truss
due to elevating the tie from a horizontal line.
Examples are given to show that when the elevation of
the tie equals i of the whole height, the vertical strain there-
by induced is equal to a weight which equals \ of half the
whole load ; and that when the elevation equals half the
whole height, the vertical strain is equal to half the whole
load. This is the strain in the vertical rod at middle. The
strains in the rafters and inclined ties are proportionately
increased.
213. — Planning a Roof. — In designing a roof for a build-
ing, the first point requiring attention is the location of the
trusses. These should be so placed as to secure solid bear-
ings upon the walls ; care being taken not to place either of
the trusses over an opening, such as those for windows or
doors, in the wall below. Ordinarily, trusses are placed so
as to be centrally over the piers between the windows ; the
number of windows consequently ruling in determining the
number of trusses and their distances from centres. This
distance should be from ten to twenty feet ; fifteen feet apart
being a suitable medium distance. The farther apart the
trusses are placed, the more they will have to carry ; not
only in having a larger surface to support, but also in that
the roof-timbers will be heavier ; for the size and weight, of
the roof-beams will increase with the span over which they
have to reach.
In the roof-covering itself, the roof-planking may be laid
upon jack-rafters, carried by purlins supported by the
trusses ; or upon roof-beams laid directly upon the back of
the principal rafters in the trusses. In either case, proper
LOAD UPON ROOFS. 189
struts should be provided, and set at proper intervals to re-
sist the bending of the rafter. Jn case purlins are used, one of
these struts should be placed at the location of each purlin.
The number of these points of support rules largely in
determining the design for the truss, thus :
For a short span, where the rafter will not require sup-
port at an intermediate point, Fig. 59 or 64 will be proper.
For a span in which the rafter requires supporting at one
intermediate point, take Fig. 60, 65, or 66.
For a span with two intermediate points of support for
the rafter, take Fig. 61 or 67.
For a span with three intermediate points, take Fig. 63.
Generally, it is found convenient to locate these points of
support at nine to twelve feet apart. They should be suffi-
ciently close to make it certain that the rafter will not be sub-
ject to the possibility of bending.
214.— Load upon Roof-Tru§§.— In constructing the force
diagram for any truss, it is requisite to determine the points
of the truss which are to serve as points of support (see
Figs. 70, 72, etc.), and to ascertain the amount of strain, or
loading, which will occur at every such point.
The points of support along the rafters will be required
to sustain the roofing timbers, the planking, the slating, the
snow, and the force of the wind. The points along the tie-
beam will have to sustain the weight of the ceiling and the
flooring of a loft within the roof, if there be one, together
with the loading upon this floor. The weight of the truss
itself must be added to the weight of roof and ceiling.
215,-lLoacl on Roof per Superficial Foot — In any im-
portant work, each of the items in Art. 214 should be care-
fully estimated, in making up the load to be carried. For
ordinary roofs, the weights may be taken per foot superficial,
as follows :
Slate, about 7-0 pounds.
Roof-plank,
Roof-beams or jack-rafters, 2 - 3
In all, I2 pounds.
CONSTRUCTION.
This is for the superficial foot of the inclined roof. For the
foot horizontal, the augmentation of load due to the angle of
the roof will be in proportion to its steepness. In ordinary
cases, the twelve pounds of the inclined surface will not be
far from fifteen pounds upon the horizontal foot.
For the roof-load we may take as follows :
Roofing, about 15 pounds.
Roof-truss, " 5
Snow, " 20 "
Wind, " 10 «.
Total on roof, 50 pounds
per square foot horizontal.
This estimate is for a roof of moderate inclination, say
one in which the height does not exceed J of the span.
Upon a steeper roof the snow would not gather so heavily,
but the wind, on the contrary, would exert a greater force.
Again, the wind acting on one side of a roof may drift the
snow from that side, and perhaps add it to that already
lodged upon the opposite side. These two, the wind and
the snow, are compensating forces. The action of the snow
is vertical : that of the wind is horizontal, or nearly so. The
power of the wind in this latitude is not more than thirty
pounds upon a superficial foot of a vertical surface ; except,
perhaps, on elevated places, as mountain-tops for example,
where it should be taken as high as fifty pounds per foot of
vertical surface.
216. — Load upon Tic-Beam. — The load upon the tie-
beam must of course be estimated according to the require-
ments of each case. If the timber is to be exposed to view,
the load to be carried will be that only of the tie-beam and
the timber struts resting upon it. If there is to be a ceiling
attached to the tie-beam, the weight to be added will be in
accordance with the material composing the ceiling. If of
wood, it need not weigh more than two or three pounds per
foot. If of lath and plaster, it will weigh about nine pounds ;
and if of iron, from ten to fifteen pounds, according to the
WEIGHT UPON ROOFS, IN DETAIL. 191
thickness of the metal. Again, if there is to be a loft in the
roof, the requisite flooring may be taken at five pounds, and
the load upon the floor at from twenty-five to seventy
pounds, according to the purpose for which it is to be used.
217 — Roof- Weights in B»eta51.— The load to be sustained
by a roof-truss has been referred to in the previous three
articles in general terms. It will now be treated more in
detail. But first a few words regarding fehe slope of the
roof. In a severe climate, roofs ought to be constructed
steeper than in a milder one, in order that snow may have a
tendency to slide off before it becomes of sufficient weight
to endanger the safety of the roof. In selecting the material
with which -the roof is to be covered, regard should be had
to the requirements of the inclination : slate and shingles
cannot be used safely on roofs of small rise. The smallest
inclination of the various kinds of covering is here given,
together with the weight per superficial foot of each.
MATERIAL.
Least Inclination.
Weight upon a
square foot.
Tin ;
F
se i inch to
a f c
or
1 to ii 1
3S.
CopDcr
i "
i to i^
v^uppci
Lead .... . .
2 inches
4 to 7
<
Zinc
3 "
i£ to 2
M
Short pine shingles
5 "
l£ tO 2
w
Long cypress shingles . .
' 6 "
2 to 3
«
Slate
6 "
5 to 9
<
The weight of the covering as here estimated includes
the weight of whatever is used to fix it in place, such as
nails, etc. The weight of that which the covering is laid
upon, such as plank, boards, or lath, is not included. The
weight of plank is about 3 pounds per foot superficial ; of
boards, 2 pounds ; and lath, about half a pound.
Generally, for a slate roof, the weight of the covering, in-
cluding plank and jack-rafters, amounts to about 12 pounds,
as stated in Art. 215 ; but in every case, the weight of each
article of the covering should be estimated, and the full load
ascertained by summing up these weights.
1 92 CONSTRUCTION.
218. — Load per Foot Horizontal. — The weight ot the
covering as referred to in the last article is the weight per
foot on the inclined surface ; but it is desirable to know how
much per foot, measured horizontally, this is equal to. The
horizontal measure of one foot of the inclined surface is
equal to the cosine of the angle of inclination. Then, to ob-
tain the inclined measure corresponding to one foot horizon-
tal, we have —
cos. : I : : p : C =
cos.
where / represents the pressure on a foot of the inclined
surface, and C the weight of so much of the inclined cover-
ing as corresponds to one foot horizontal. The cosine of an
angle is equal to the base of the right-angled triangle divided
by the hypothenuse (see Trigonometrical Terms, Art. 474),
which in this case is half the span divided by the length of
the rafter, or — -. , where s is the span, and / the length of the
rafter. Hence, the load per foot horizontal equals —
p p _2 Ip
^c^sT^T1 ~T~ (92.)
2/
or, twice the pressure per foot of inclined surface multiplied
by the length of the rafter and divided by the span, both in
feet, will give the weight per foot measured horizontally.
219. — WeigBit of Tru§§. — The weight of the framed truss
will be in proportion to the load and to the span. This, for
the weight upon a foot horizontal, will about equal—
T — 0-077 Cs\
which equals the weight in pounds per foot horizontal to be
allowed for a wooden truss with iron suspension-rods and a
horizontal tie-beam, near enough for the requirements of our
present purpose ; where s- equals the length or span -of the
EFFECT OF WIND ON ROOFS. 193
truss, and C the weight per foot horizontal of the roof cover-
ing, as in equation (92.). Substituting for Cits value, as in
(92.), we have —
T= 0-0077^;
or —
T = 0-0154 lp\ (93.)
which equals the weight in pounds per foot horizontal to
be allowed for the truss.
220. — Weight of Snow on Rooffc. — The weight of snow
will be in proportion to the depth it acquires, which will be
in proportion to the rigor of the climate of the place where
the building is to be erected. Upon roofs of ordinary incli-
nation, snow, if deposited in the absence, of wind, will not
slide off ; at least until after it has acquired some depth, and
then the tendency to slide will be in proportion to the angle
of inclination. The weight of snow may be taken, therefore,
at its weight per cubic foot (8 pounds) multiplied by the
depth it is usual for it to acquire. This, in the latitude of
New York, may be taken at about 2-J- feet. Its. weight
would, therefore, be 20 pounds per foot superficial, meas-
ured horizontally.
221. — Effect of Wind on Roof*. — The direction of wind
is horizontal, or nearly so, when unobstructed. Precipitous
mountains or tall buildings deflect the wind considerably
from its usual horizontal direction. Its power usually does
not exceed 30 pounds per superficial foot except on ele-
vated places, where it sometimes reaches 50 pounds or more.
This is the pressure upon a vertical surface ; roofs, however,
generally present to the wind an inclined surface. The ef-
fect of a horizontal force on an inclined surface is in pro-
portion to the sine of the angle of inclination ; the direction
of this effect being at right angles to the inclined surface.
The force thus acting may be resolved into forces acting in
two directions— the one horizontal, the other vertical ; the
former tending, in the case of a roof, to thrust aside the walls
194
CONSTRUCTION.
on which the roof rests, and the latter acting directly on the
materials of which the roof is constructed — this latter force
being in proportion to the sine of the angle of inclination
multiplied by the cosine. This will be made clear by the
following explanation. Re-
ferring to Fig. 83, let D KE
be the angle of inclination
of the roof, D E being equal
to one foot. Bisect DK at
A ; draw A L parallel with
FIG. 83. EK\ make A L equal to the
horizontal pressure of the wind upon one foot superficial of
a vertical plane. Draw A C perpendicular to D K, and LF
parallel with A C from F draw FC parallel with EK\ draw
A B parallel with D E. The sides of the triangle LA F rep-
resent the three several forces in equilibrium : LA is the
force of the wind ; L F is the pressure upon the roof ; and
A F is the force with which the wind moves on up the roof
towards D. Now, to find the relation of the force of the
wind to the strain produced by it in the direction A C, we
have —
rad.
rad. : sin. \\FC\AC\
F C = LA; therefore—
: sin. : : L A : A C — L A sin. ;
AC = Fsin.-t
or, the strain perpendicular to the surface of the roof equals
the force of the wind multiplied by the sine of the angle of
inclination. When A C represents this strain, then, of the
two forces referred to above, B C represents the horizontal
force, and A B the vertical force. To obtain this last force,
we have —
rad. : cos. \\AC\AB.
Putting for A C its value as above, we have —
rad. : cos. : : /^sin. : A B = F sin. cos.;
V = F sin. cos. ;
FORCE OF THE WIND. 195
or, the vertical effect is equal to the product of the force of
the wind upon a superficial foot into the sine and the cosine
of the angle of inclination. This result is that which is due
to the pressure of the wind upon so much of the inclined
surface as is covered by one square foot of a vertical sur-
face. The wind, acting horizontally through one foot super-
ficial of vertical section, acts on an area of inclined surface
equal to the reciprocal of the sine of inclination, and the
horizontal measurement of this inclined surface is equal to
the cosine of the angle of inclination divided by the sine.
This may be illustrated from Fig. 83, thus—
sin. : rad. \\DE\DK.
D E equals I foot ; therefore — • .+•.
sin. : rad. : : I : D K = — !— ;
sin.
or, the surface acted upon by one square foot of sectional
area equals the reciprocal of the sine of the angle of incli-
nation. Again, the horizontal measure of this inclined sur-
face may be obtained thus—
cos.
sin. : cos. : : D E : K E = — — ;
sin.
or, KE, the horizontal measurement, equals the cosine of the
angle of inclination divided by the sine.
In tile figure, make K G equal to one foot ; then we
have —
K E : KG : : V '. W\
in which V, as above, represents the vertical pressure due to
the wind acting upon the surface KD, and W the vertical
pressure due to the wind acting upon the surface KH, or
so much as covers KG, one foot horizontal.
Now we have, as above, K E equal to — -— , K. G = i, and
196 CONSTRUCTION.
V — F sin. cos. Substituting these values, we have, instead
of the above proportion-
cos. _ .
— — : i : : F sin. cos. : W\
sin.
from which —
'f±? (94.)
sn.
or, the vertical effect of the wind upon so much of the roof
as covers each square foot horizontal, is equal to the pro-
duct of the force of the wind per square foot into the square
of the sine of the angle of inclination.
Example. — When the force of the wind upon a square
foot of vertical surface is 30 pounds, what will be the verti-
cal effect per square foot horizontal upon a roof the inclina-
tion of which is 26° 33', or 6 inches to the foot?
Here we have F = 30, and the sine of 26° 33' is 0-44698 ;
therefore—
W— 30 x 0-44698 2 = 5-9937-
This is conveniently solved by logarithms ; thus—
log. 30 = 1^-4771213
0-44698 = 9-6502868
0-44698 = 9^-6502868
5-9937 = 0-7776949
or, the vertical effect is (5 -9937, or) 6 pounds.
The form of equation (94.) may be changed ; for, in a right-
angled triangle, the sine of the angle at the base is equal to
the perpendicular divided by the hypothenuse ; which, in
the case of a roof, is the height divided by the length of the
rafter; or —
height h
Sme = TaTteF = 2
LOAD UPON ROOFS. 197
Therefore, equation (94.) may be changed to —
(950
or, the vertical effect upon each square foot of a roof is equal
to the product of the force of the wind per foot into the
square of the height of the roof at the ridge, divided by the
square of the length of the rafter (the height and length both
in feet.)
Example. — When the force of the wind is 30 pounds, the
height of the roof 10 feet, and the length of the rafter 22-36
feet, what will be the vertical effect of the wind ? Here we
have F ~ 30, h = 10, and / = 22-36 ; and—
222. — Total Load per Foot Horizontal. — The various
items comprising the total load upon a roof are the cover-
ing, the truss, the wind, snow, the plastering or other kind
of ceiling, and the load which may be deposited upon a floor
formed in the roof ; or, the total load —
M= C+T + W+S + P+L.
The value per foot horizontal for these has been found as
follows: C=^; T= 0-0154 //; W=F^. For 5 the
value must be taken according to circumstances, as in Art.
220. So, also, the value of P and L are to be assigned as
required for each particular case, as in Art. 216. The total
load, therefore, with these substitutions, will be—
M =
which reduces to —
M = lp (- + 0-0154) + F^t \-S + P+L; (96.)
* S / l>
198 CONSTRUCTION.
in which / is the length of the rafter ; / is the weight of the
covering per foot superficial, including the roof boards or
slats, the jack-rafters, etc. ; s is the span of the roof ; ~h is the
vertical height above a horizontal line passing through the
feet of the rafters ; F is the force of the wind per square foot
against a vertical surface ; 5 is the weight of snow per
square foot horizontal ; P is the weight per superficial foot
of the ceiling at the tie-beam ; and L, the load per superficial
foot in the roof, including weight of flooring and floor-
timbers. The dimensions, s, /, and /i, are each in feet ; the
weight of /, F, 5, P, and L are each in pounds. The value
of/ is for a square foot of the inclined surface.
223. — Strain§ in Roof-Timbers Computed.— The graphic
method of obtaining the strains, as shown in Arts. 205 to 211,
is, for its conciseness and simplicity, to be preferred to any
other method ; yet, on some accounts, the method of obtain-
ing the strains by the parallelogram of forces and by arith-
metical computations will be found useful, and will now be
referred to.
By the parallelogram of forces, the weight of the roof is
in proportion to the oblique thrust or pressure in the axis of
the rafter as twice the height of the roof is to the length of
the rafter ; or —
R: F:: 2 A: /;
or, transposing —
2&:i::R:Y=j£; (97-)
where F equals the pressure in the axis of the rafter, and R
the weight of one truss and its load. Again, the weight of
the roof is in proportion to the horizontal thrust in the tie-
beam as twice the height of the roof is to half the span ; or —
2
or, transposing —
2/i:S-::R:H=~', (98.)
THE STRAINS SHOWN GEOMETRICALLY. 199
where H equals the horizontal thrust in the tie-beam. To
obtain R, the weight of the roof, multiply M, the load per
foot, as in equation (96.), by s, the span, and by c, the dis-
tance from centres at which the trusses are placed ; or
R = M c s.
With this value of R substituted for it, we have —
K= — — '
and —
TT M c s* , .
H = T— ; (loo.)
4 It
in which F equals the strain in the axis of the rafter, and H
the strain in the tie-beam. These are the greatest strains
in the rafter and tie-beam. At certain parts of these pieces
the strains are less, as will be shown in the next article.
224. — Strains in Roof-Timber§ Shown Geometrically. —
The pressure in each timber may be obtained as shown in
Fig. 84, where A B represents the axis of the tie-beam, A C
the axis of the rafter, D E and F B the axes of the braces,
and DG, FE, and C B the axes of the suspension-rods. In
this design for a truss, the distance A B is divided into three
equal parts, and the rods located at the two points of division,
G and E. By this arrangement the rafter A £7 is supported at
equidistant points, D and F. The point D supports the rafter
for a distance extending half-way to A and half-way to F, and
the point F sustains half-way to D and half-way to C. Also,
the point C sustains half-way to F, and, on the other rafter,
half-way to the corresponding point to F. And because these
points of support are located at equal distances apart, there-
fore the load on each is the same in amount. On D G make
Da equal by any decimally divided scale to the number of
hundreds of pounds in the load on D, and draw the parallel-
ogram abDc. Then, by the same scale, Db represents
(Art. 71) the pressure in the axis of the rafter by the load at
2OO
CONSTRUCTION,
FIG. 84.
STRAINS IN A. TRUSS. 2OI
D\ also, DC the pressure in the brace D E. Draw cd hori-
zontal ; then D d is the vertical pressure exerted by the brace
D E at E. The point F sustains, besides the common load
represented by D a, also the vertical pressure exerted by the
brace D E ; therefore, make Fe equal to the sum of D a and
Dd, and draw the parallelogram F gef. Then Fg, meas-
ured by the scale, is the pressure in the axis of the rafter
caused by the load at F, and F f is the load in the axis of the
brace FB. Draw fh horizontal ; then Fh is the vertical
pressure exerted by the brace Ffiat B. The point C, besides
the common load represented by D a, sustains the vertical
pressure Fh caused by, the brace FB, and a like amount
from the corresponding brace on the opposite side. There-
fore, make Cj equal to the sum of Da and twice Fit, and
draw jk parallel to the opposite rafter. Then Ck is the
pressure in the axis of the rafter at C. This is not the only
pressure in the rafter, although it is the total pressure at its
head C. At the point F, besides the pressure C k, there is
F g. At the point D, besides these two pressures, there is
the pressure D b. At the foot, at A, there is still an addi-
tional pressure ; for while the point D sustains the load half-
way to F and half-way to A, the point A sustains the load
half-way to D. This load is, in this case, just half the load
at D. Therefore draw A m vertical, and equal, by the scale,
to half of Da. Extend CA to/; draw ml horizontal.
Then A I is the pressure in the rafter at A caused by the
weight of the roof from A half-way to D. Now the total of
the pressures in the rafter is equal to the sum of A 1+ D b +
Fg added to C k. Therefore make kn equal to the sum of
A l+Db + F g, and draw no parallel with the opposite raf-
ter, and nj horizontal. Then Co, measured by the same
scale, will be found equal to the total weight of the roof on
both sides of C B. Since Da represents s, the portion of the
weight borne by the point D, therefore Co, representing the
whole weight of the roof, should equal six times Da, as it
does, because D supports just one sixth of the whole load.
Since C n is the total oblique thrust in the axis of the rafter
at its foot, therefore nj is the horizontal thrust in the tie-
beam at A.
2O2 CONSTRUCTION.
225.— Application of the Geometrical System of Strains.—
The strains in a roof-truss can be ascertained geometrically,
as shown in Art. 224. To make a practical application of
the results, in any particular case, it is requisite first to as-
certain the load at the head of each brace, as represented by
the line D a, Fig. 84. The load corresponding to any part
of the roof is equal to the product of the superficial area of
that particular part (measured horizontally) multiplied by
the weight per square foot of the roof. Or, when M equals
the weight per square foot, c the distance from centres at
which the trusses are placed, and n the horizontal distance
between the heads of the braces, then the total load at the
head of a brace is represented by—
N—Mcn. (101.)
The value of M is given in general terms in equation (96.).
To show its actual value, let it be required to find the weight
per square foot upon a root 52 feet span and 13 feet high at
middle ; or (Fig. 84), where A B equals half the space, or 26
feet, and C B 13 feet, then ^4 C, the length of the rafter, will
be 26-069, nearly. And where the weight of covering per
square foot, on the inclination, is 12 pounds, the force of the
wind against a vertical plane is 30 pounds ; the weight of
snow per foot horizontal is 20 pounds ; the weight of the
plastering forming the ceiling at the tie-beam is 9 pounds ;
and the load in the roof is nothing ; — with these quantities
substituted, equation (96.) becomes—
M — 2Q'o6gx 12 (-- 4- 0-0154) 4- 30 — 1^ — + 20 49 4-0 ;
\52 / 29-069'
M = (29-069 X 12 X 0-05386) 4 (30 x 0-2) + 20 4- 9;
M = 18-788 4-6 + 29 = 53-788;
or, say, 53-8 pounds. Then if <r, the distance from centres
between trusses, is 10 feet, and ;/, the distance between
braces, is one third of A B, Fig. 84, or 2/ = 8f , the total load
at the head of a brace will be, as per equation (101.) —
N— 53-8 x 10 x 8f = 4663;
TABLE OF STRAINS. 203
or, say, 4650 pounds. Now, by any decimally divided scale,
make D a, Fig. 84, equal to 46^ parts of the scale ; this being
the number of hundreds of pounds contained in the weight
at D, as above. Then, by the same scale, the several lines
in the figure drawn as before shown will be found to repre-
sent respectively the weights here set opposite to them, as
follows :
D d — da — he — 23^, and represents 2325 pounds;
" 4650
" 5200 "
F e = D a + D d — 69$ " 6975
Ff = 65| « 6575
Cj — 3 D a — 139-1- " 13950
CK = $ D b = \$6 " 15600
C n = C k +- Fg+ D b + A 1= 312 " 31200 "
Cn^Ck + $Db=6 Db=2 Ck
= 312 " 31200
Nj — C o — 6 D a — 6^ 4.6% — 279 " 27900 *'
It should be observed here that the equality of the lines nj
and Co is a coincidence dependent upon the relation which
in this particular case the line CB happens to bear to the
line A B ; A B being equal to twice C B. And so of some
other lines in the figure. If the inclination of the roof were
made greater or less, the equality of the lines referred to
would disappear. It should also fye observed that the strains
above found are not quite exact ; they are, however, correct
to within a fraction of a hundred pounds, which is a suffi-
ciently near approximation for the purpose intended. From
the results obtained above, we ascertain that the strain in
the rafter, from F to C, is represented by C K, and is equal
to 15,600 pounds ; while the strain at the foot of the rafter,
from A to D, is represented by C n, and equals 3 1,200 pounds,
or double that which is at the head of the rafter. We ascer-
tain, also, that the maximum strain in the tie-beam, repre-
sented by 11 j, is 27,900 pounds; that that in the brace D E,
represented by DC, is 5200 pounds; and that that in the
brace F B, represented by Ff, is 6575 pounds. The strain
204 CONSTRUCTION.
in the vertical rod D G is theoretically nothing. There is,
however, a small strain in it, for it has to carry a part of the
tie-beam and so much of the ceiling as depends for support
upon that part. But the manner of locating the weights,
adopted in this article, does not recognize any load located
at the point G. This is an objection to this system, but it
is not material.
For a recognition of weights at the tie-beam, see Arts.
205 to 211. The load at G may be found by obtaining the
product of the surface carried into the weight per foot of
the ceiling; or, say, 10 c n = 10 x lox 8f = 867 pounds.
The load to be carried by the rod F E is shown at D d= he,
which above is found to be 2325 pounds. To this is to be
added 867 pounds for the ceiling at E, as before found for the
ceiling at G\ or, together, 3192 pounds. The central rod
C B has to carrv the two loads brought to B by the two
braces footed there ; and also the weight of the ceiling sup-
ported by B. The vertical strain from the brace F B is rep-
resented at Fh, and equals 4650 pounds ; therefore, the
total load on CB is 4650 + 4650 -f 867 = 10,167 pounds.
226. — Roof-Timber* : the Tie-Beam. — The roof-timbers
comprised in the truss shown in Fig. 84 are the rafters,
tie-beam, two braces, and three rods. Of these, taking first
the tie-beam, we have a piece subject to tension and some-
times to cross-strain (see Art. 682, Transverse Strains}. In
this case the tensile strain "only need be considered. For
this a rule is given in Art. 117. In this rule, if the factor of
safety be taken at 20, the result will be sufficiently large to
allow for necessary cuttings at the joints. Therefore, if the
beam be of Georgia pine, equation (16.), Art. 117, becomes —
__ 27900x20 _
~~ ~
or, say, 35 inches. This is ample to resist the tensile strain ;
but, to resist the transverse strains to which such a long
piece of timber is subjected in the hands of the workman,
it would be proper to make it, say, 6x9.
STRAIN UPON THE RAFTER. 205
227. — The Rafter. — A rafter, like a post, is subject to a
compressive force, and is liable to fail in three ways, name-
ly : by flexure, by being crushed, or by crushing the material
against which it presses. To render it entirely safe, there-
fore, it is requisite to ascertain the requirements for resisting
failure in each of these three ways.
Of these it will be convenient to consider, first, that of
the liability to being crushed. The rule for this is found in
Art. 107. Let the rafter be of Georgia pine, then the vajue
of C, Table I., will be 9500. The strain in the rafter (Art.
22$) is 31,200 pounds. Now, taking the value of a, the fac-
tor of safety, at 10, we have, by Rule VI. (Art. 107.) —
31200 x 10
A = 9500 = 32'737:
or, 33 inches area of cross-section. This is the size of the
rafter at its smallest section ; for example, at any one of the
joints where it is customary to reduce the area by cutting
for the struts and rods.
Again : Let the liability of the rafter to flexure be now
considered. For this we have a rule in Art. 114. The
length of the rafter between unsupported points is nearly 9$
feet, or 9! x 12 = 1 16 inches. Let the thickness of the rafter
be taken at 6 inches. Then, by Rule XI. (Art. 114), we
have —
+ fyer3) _ 31200 x 10(14- f x -00109 x r*).
c r~ 9500 x 6
/ 116
r=7--g-=i9i; 19* * : 373-8-
Then, f x -00109 x 373-8 = 0-611127
adding unity = I •
1-611127
Substituting this, we have—
31200 x IPX 1-611127 _. 50267 1 '624 _
" 9500 x~6~ "57000"
206 CONSTRUCTION.
or, to resist flexure the breadth is required to be 8-82, or,
say, 9 inches ; or, the rafter is to be 6 xg inches at the foot.
The strain in the rafter at the upper end is only half that at
the foot ; the area of cross-section, therefore, at the head
need not be more than half that which is required at the
foot ; but it is usual to make it there about f of the size at the
foot. In this case it would be, therefore, 6x6 inches at the
upper end.
Lastly, the requirement to resist crushing the surfaces
against which the rafter presses is to be considered.
The fibres of timber yield much more readily when
pressed together by a force acting at right angles to the di-
rection of their length than when it acts in a line with their
length.
The value of timber subjected to pressure in these two
ways is shown in Arts. 94, 98. In Table I., the value per
square inch of the first stated resistance is expressed by P,
and the ultimate resistance of the other by — . The value
of timber per square inch to safely resist crushing may be ex-
pressed by — , in which a is the factor of safety. Timber
pressed in an oblique direction will resist a force exceeding
£
that expressed by P, and less than that expressed by -— .
ct
When the angle of inclination at which the force acts is just
45°, then the force will be an average between P and — .
And for any angle of inclination, the force will vary inverse-
ly as the angle ; approaching P as the angle is enlarged,
but approaching - - as the angle is diminished. It will be
equal to -- when the angle becomes zero, and equal P
when the angle becomes 90°. The resistance of timber per
square inch to an oblique force is therefore expressed by—
M =
RESISTANCE OF SURFACES. 2O?
where A° equals the complement of the angle of inclination.
In a roof, A° is the acute angle formed by the rafter with
a vertical line. If no convenient instrument be at hand to
measure the angle, describe an arc upon the plan of the
truss — thus : with C B (Fig. 84) for radius, describe the arc
B g, and get the length of this arc in feet by stepping it off
with a pair of dividers. Then —
where k equals the length of the arc, and h equals B C, the
height of the roof. Therefore — •
M =
equals the value of timber per square inch in a tie-beam, C
and P being obtained from Table I., Art. 94. When C for
the kind of wood in the tie-beam exceeds C set opposite the
kind of wood in the rafter, then the latter is to be used in
the rules instead of the former.
The value of M, equation (103.),. is the resistance per
square inch of the surface pressed at the foot of the rafter.
The resistance of the entire surface will therefore be MA,
where A equals the area of the joint. Then, when the re-
sistance equals the strain, we will have —
from which we have —
in which 5 is the strain to be resisted.
Now, the end of the rafter must be of sufficient size to
afford a joint the area of which will not be less than that
expressed by A in equation (104.).
For example, the strain to which the rafter, Fig. 84, is
subject at its foot is ascertained to be (Art. 225) 31,200 pounds.
For Georgia pine, the material of the tie-beam, P = 900
(Art. 94, Table /.), and ^ = 9500.
208 CONSTRUCTION.
The length of the arc Bg is about 14-4 feet; the height
B Cis 13 feet. Let a, the factor of safety, be taken at 10,
then we have (104.) —
31200
900 + (o-63|
=
x 50) -
or, the superficial area of the bearing at the joint required
to prevent crushing the tie-beam is 33^- inches.
The results of the computations show that the rafter is
required to be 6 inches thick, 9 inches wide at the foot, and
6 inches wide at the top. It is also ascertained that, in cut-
ting for the bearing for the struts and boring for the sus-
pension-rods, it is required that there shall be at least 33
inches area of cross-section left intact ; and, farther, that the
area of the surface of the joint against the tie-beam should
not be less than 33^ inches.
228. — The Braces. — Each brace is subject to compres-
sion, and is liable to fail if too small, in the same manner as the
rafter. Its size is to be ascertained, therefore, in the manner
described for the rafter ; which need not be here repeated,
except, perhaps, as to the liability to fail by flexure ; for in this
case we have the breadth given, and need to find the thick-
ness. The breadth of the brace is fixed by the thickness of
the rafter, for it is usual to have the two pieces flush with
each other. Rule XI. (Art. 114) is to be used, but with this
difference, namely : instead of the thickness, use the breadth
as one of the factors in the divisor. Thus —
(105.)
In working this rule, it is required, in order to get the
value of r, the ratio between the height and thickness, to
assume the thickness before it is ascertained ; and after com-
putation, if the result shows that the assumed value was not
a near approximation, a second trial will have to be made.
Usually the first trial will be sufficient.
STRAIN UPON BRACES. 209
For example, the brace D E is about 9$ feet or 1 16 inches
long. As the strain in it is only 5200 pounds, the thickness
will probably be not over 3 inches. Assuming it at this, we
have r = -= -*-J-& = 38$ ; the square of whichxis about
Therefore, we have- 4 N I VE R S I T Y
|-XO-OOI09XI495=2
add unity = i.
^444S
The equation reduces, therefore, to this —
, = 5200 x 10XV4445
9500 x 6
or, the required thickness of the brace is 3^- inches, or the
brace should be, say, 3 J x 6 inches. In this case the result is
so near the assumed value, a second trial. is not needed.
For the second brace, we have the length equal to about
\2\ feet or 147 inches; and the strain equal to 6575 pounds
(Art. 22$). The ratio, therefore, may be obtained by assum-
ing the thickness, say, at 4. With this, we have —
i^— 36-75 ; the square of which is 1350^'
With this value of r2—
| x « 00109 x 1350^ = 2- 2081
add unity = I.
3-2081
Then —
6575 x lox 3-2081 .
t=.^L2 — _^ - = 3.7006.
9500x6
Comparing this result with the assumed value of / = 4»
we find the difference so great as to require a second trial.
As the value of r was taken too low, the result obtained is
correspondingly low. The true value is somewhere between
3 - 7 and 4. Assume it now, say, at 3 - 9. With this value, we
have —
r = -== — = 37-692 ; the square of which is 1420-7.
210 CONSTRUCTION.
With this value of r*—
| x -00109 x 1420-7 = 2-32282
add unity = i •
3-32282
Then —
, _ 6575x10x3-32282 _
9500x6
This result is a trifle less than the assumed value, 3-9. The
true value is between these, and probably is about 3-86.
This is quite near enough for use. This brace, therefore, is
required to be 3 - 86 x 6 inches, or, say, 4x6 inches.
229. — The Su§pension-Rods.— These are usually made of
wrought iron. This metal, when of excellent quality, may
be safely trusted with 12,000 pounds per inch sectional area.
But it is usual, for good work, to compute the area at only
9000 pounds per inch, and, as ordinarily made, these rods
ought not to be loaded with more than 7000 pounds. The
strain divided by this value per inch of the metal will give
the sectional area of cross-section. For example, the strain in
the rod D G, Fig. 84, is 867 pounds (Art. 225); therefore—
867
or, the sectional area required is only an eighth of an inch.
By reference to the table of areas of circles in the Appen-
dix, the diameter of a rod containing the required area, as
above, will be found to be a little less than half an inch. A
rod half an inch in diameter will therefore be of ample
strength. For appearance's sake, however, no rod in a truss
should be less than f of an inch in diameter.
•The rod FE has to resist a strain of 3192 pounds. For
this, then, we have —
A reference to the table of areas shows that a rod contain-
ROOF-BEAMS. 211
ing this area would be a little more than J of an inch in di-
ameter ; it would be of ample strength, say, at £ of an inch
in diameter.
The rod C B, at the centre, has to carry a strain of 10,167
pounds. For this, then,- we have —
10167
A reference to the table of areas shows that this rod should
be i£ inches in diameter.
230. — Roof-Beams, Jack-Rafters, and Purlins. — These
timbers are subject to loads nearly uniformly distributed,
and their dimensions may be obtained by Rule XXX., equa-
tion (35.), Art. 140. In this equation, U=cfl(Art. 152).
Substituting this value for U, and r I for tf, equation (35.) be-
comes —
and putting for r the rate of deflection, .04, we have —
a formula convenient for roof-timbers.
Example. — In a roof where the roofing is to be supported
on white-pine roof-beams 10 feet long, placed 2\ feet from
centres, and where the load per foot superficial is to be 40
pounds, including wind and snow : what should be the di-
mensions of the roof-beams? By equation (106.)—
bd =
Now if b, the breadth, be fixed, say, at 3, then—
d = 5-64 nearly.
212
CONSTRUCTION.
The roof-beams, therefore, require to be 3 x 5$, or, say, 3x6.
All pieces of timber subject to cross-strains will sustain
safely much greater strains when extended in one piece over
two, three, or more distances between bearings ; therefore,
roof-beams, jack-rafters, and purlins should, if possible, be
made in as long lengths as practicable ; the roof-beams and
purlins laid on, not framed into, the principal rafters, and
extended over at least two spaces, the joints alternating on
the trusses ; and likewise the jack-rafters laid on the purlins
in long lengths.
231. — Five Examples of Rooft: are shown at Figs. 85, 86,
87, 88, and 89. In Fig. 85, a is an iron suspension-rod, by b are
braces. In Fig. 86, a, a, and
b are iron rods, and d, d, c, c
are braces. In Fig. 87, a, b
are iron rods, d, d braces, and
c the straining beam. In
FIG. 85. Fig. 88, a, a, b, b are iron rods,
, e> d, d are braces, and c is a straining beam. In Fig. 89, pur-
out
S<rft. I
lins are located at PP, etc. ; the inclined beam that lies upon
them is the jack-rafter; the post at the ridge is the king-
TRUSS WITH BUILT-RIB.
2I3
rTi
?k
o a
post, the others are queen-posts. In this design the tie-beam
is increased in height along the middle by a strengthening
piece (Art. 163), for the purpose of sustaining additional
weight placed in the room form-
ed in the truss (Art. 216).
Fig. 90 shows a method of
constructing a truss having a
built-rib in the place of prin-
cipal rafters. The proper form
for the curve is that of the par-
abola (Art. 560). This curve,
when as flat as is described in
the figure, approximates so close-
ly to that of the circle that the
latter may be used in its stead.
The height, a b, is just half of
a c, the curve to pass through
the middle of the rib. The rib
is composed of two series of
abutting pieces, bolted together, oo ^
These pieces should be as long
as the dimensions of the timber
will admit, in order that there
may be but few joints. The sus-
pending pieces are in halves,
notched and bolted to the tie-
beam and rib, and a purlin is
framed upon the upper end of
each. A truss of this construc-
tion needs, for ordinary roofs,
no diagonal braces between the
suspending pieces, but if extra
strength is required the braces
may be added. The best place \~^~.\
for the suspending pieces is at
the joints of the rib. A rib of this kind will be sufficiently
strong if the area of its section contain about one fourth
more timber than is required for that of a rafter for a roof
of the same size. The proportion of the depth to the thick-
ness should be about as 10 to 7.
214
CONSTRUCTION.
232. — Roof-Tru§s with Elevated Tic-Beam. — Designs
such as are shown in Fig. 91 have the tie elevated for the ac-
commodation of an arch in the ceiling. This and all similar
designs are seriously objectionable, and should always be
avoided ; as the smail height gained by the omission of the
tie-beam can never compensate for the powerful lateral
strains which are exerted by the oblique position of the
supports, tending to separate the walls. Where an arch is
required in the ceiling, the best plan is to carry up the
walls as hi^h as the top of the arch. Then, by using a
horizontal tie-beam, the oblique strains will be entirely re-
HIP-ROOFS.
215
moved. It is well known that many a public building has
been all but ruined by the settling of the roof, consequent
upon a defective plan in the formation of the truss in this
respect. It is very necessary, therefore, that the horizontal
FIG. 91.
tie-beam be used, except where the walls are made so strong
and firm by buttresses, or other support, as to prevent a
possibility of their separating. (See Art. 212.)
233.— Hip-Roofs: Lines and Bevils.— The lines a b and
be in Fig. 92, represent the walls at the angle of a building;
b e is the seat of the hip-rafter, and gfoi a jack or cnppl(
rafter. Draw e h at right angles to be, and make it equal
2l6
CONSTRUCTION.
to the rise of the roof ; join b and //, and h b will be the
length of the hip-rafter. Through e draw di at right angles
to b c\ upon b, with the radius b h, describe the arc hi,
cutting di in z ; join b and i, and extend gfto meet b i in/;
then gj will be the length of the jack-rafter. The length
of each jack-rafter is found in the same manner — by extend-
ing its seat to cut the line b i. From / draw fk at right
angles tofg, also// at right angles to be\ make/£ equal
to// by the arc Ik, or make g k equal to gj by the arc jk ;
then the angle at / will be the top-bevil of the jack-rafters,
and the one at k will be the down-bevil*
234. — The Backing of the Hip-Rafter. — At any con-
venient place in be (Fig. 92), as o, draw m n at right angles to
b e ; from o, tangical to b h, describe a semicircle, cutting b e
in s ; join m and s and n and s\ then these lines will form at
s the proper angle for bevilling the top of the hip-rafter.
DOMES.f
235. — Domes. — The usual form for domes is that of the
sphere ; the base circular. When the interior dome does not
FIG. 93.
rise too high, a horizontal tie may be thrown across, by
which any degree of strength required may be obtained.
* The lengths and bevils of rafters for root-valleys can also be found by the
above process.
f See also Art. 68.
CONSTRUCTION OF DOMES.
217
- 93 shows a section, and Fig. 94 the plan, of a dome of
this kind, a b being the tie-beam in both. Two trusses of
this kind (Fig. 93), parallel to each other, are to be placed
one on each side of the opening in the top of the dome.
Upon these the whole framework is to depend for support,
FIG. 94.
u u
and their strength must be calculated accordingly. (See
Arts. 70 to 80 and 214 to 222.) If the dome is large and of
importance, two other trusses may be introduced at right
angles to the foregoing, the tie-beams being preserved in
;i*/ FIG. 95.
one continuous length by framing them high enough to pass
over the others.
236.— Ribbed Dome.— When the interior must be kept
free, then the framing may be composed of a succession of
ribs standing upon a continuous circular curb of timber, as
218
CONSTRUCTION.
seen at Figs. 95 and 96 — the latter being a plan and the former
a section. This curb must be well secured, as it serves hi
the place of a tie-beam to resist the lateral thrust of the ribs.
In small domes these ribs may be easily cut from wide
plank ; but where an extensive structure is required, they
must be built in two thicknesses so as to break joints, in the
same manner as is described for a roof at Art. 231. They
should be placed at about two feet apart at the base, and
strutted as at a in Fig. 95.
FIG. .96.
The scantling of each thickness of the rib may be as fol-
lows :
For domes of 24 feet diameter, i x 8 inches.
36 " " i£ x 10 "
2 x 13 "
a
" 60
" 90
"108
3 x 13
237. — Dome: Curve of Equilibrium. — The surfaces of a
dome may be finished to any curve that may be desired, but
the framing should be constructed of such form that the
curve of equilibrium shall be sure to pass through the middle
of the depth of the framing. The nature of this curve is
such that, if an arch or dome be constructed in accordance
with it, no one part of the structure will be less capable than
another of resisting the strains and pressures to which the
whole fabric may be exposed. The curve of equilibrium for
an arched vault or a roof, where the load is equally diffused
CURVE OF EQUILIBRIUM.
2I9
over the whole surface, is that of a. parabola (Art. 460}-, for
a dome having no lantern, tower, or cupola above it, a cubic
parabola (Fig. 97) ; and for one having a tower, etc., above it,
a curve approaching that of an hyperbola must be adopted,
as the greatest strength is required at its upper parts. If
the curve of a dome be circular (as in the vertical section,
^•95)> the pressure will have a tendency to burst the dome
outwards at about one third of its height. Therefore, when
this form is used in the construction of an extensive dome,
an iron band should be placed around the framework at
that height ; and whatever may be the form of the curve, a
band or tie of some kind is necessary around or across the
base.
0
o/
0
/
/I
1
of \
A
»
s f
FIG. 97.
If the framing be of a form less convex than the curve of
equilibrium, the weight will have a tendency to crush the
ribs inwards, but this pressure may be effectually overcome
by strutting between the ribs ; and hence it is important
that the struts be so placed as to form continuous horizontal
circles.
238. — Cubic Parabola Computed. — Let a b (Fig. 97) be
the base, and b c the height. Bisect a b at d, and divide a d
into 100 equal parts ; of these give d e 26, ef \%\,fg i&,gh
\2\, h i lof, ij 9^, and the balance, 8j, to/# ; divide be into
8 equal parts, and from the points of division draw lines
parallel to a b, to meet perpendiculars from the several points
22O
CONSTRUCTION.
of division in a b, at the points o, o, oy etc. Then a curve
traced through these points will be the one required.
239. — Small Domes over Stairways : are frequently made
elliptical in both plan and section ; and as no two of the ribs
in one quarter of the dome are alike in form, a method for
obtaining the curves may be useful.
FIG. 99,
To find the curves for the ribs of an elliptical dome, let
abed (Fig. 98) be the plan of a dome, and ef the seat of
one of the ribs, Then take c /for the transverse axis and
twice the rise, og> of the dome for the conjugate, and de-
COVERING OF DOMES.
221
scribe (according to Arts. 548, 549, etc.) the semi-ellipse
e gf, which will be the curve required for the rib e gf. The
other ribs are found in the same manner.
240. — Covering for a Spherical Dome. — To find the
shape, let^4 C^T-99) be the plan, and B the section, of a given
dome. From a draw a c at right angles to a b ; find the
stretch-out (Art. 524) of o b, and make dc equal to it; divide
the arc o b and the line d c each into a like number of equal
parts, as 5 (a large number will insure greater accuracy than
a small one) ; upon c, through the several points of division
in cd, describe the arcs o do, i e I, 2/2, etc. ; make do equal
to half the width of one of the boards, and draw o s parallel
to a c ; join s and #, and from the points of division in the arc
<?£drop perpendiculars, meeting a sinij kl\ from these
points draw z 4,/3, etc., parallel to ac ; make dotei, etc., on
the lower side of a c, equal to do,e\, etc., on the upper side ;
trace a curve through the points o, 1,2, 3, 4, c, on each side
of dc ; then o c o will be the proper shape for the board. By
dividing the circumference of the base A into equal parts,
and making the bottom, o d o, of the board of a size equal to
one of those parts, every board may be made of the same
size. In the same manner as the above, the shape of the
covering for sections of another form may be found, such as
an ogee, cove, etc.
To find the curve of the boards when laid in horizontal
courses, let A B C (Fig. 100) be the section of a given dome,
222
CONSTRUCTION.
and DB its axis. Divide B C into as many parts as there
are to be courses of boards, in the points i, 2, 3, etc. ; through
i and 2 draw a line to meet the axis extended at a\ then a
will be the centre for describing the edges of the board F.
Through 3 and 2 draw 3 b; then b will be the centre for de-
FlG. IOI.
scribing F. Through 4 and 3 draw 4^; then d will be the
centre for G. B is the centre for the arc i o. If this method
is taken to find the centres for the boards at the base of the
dome, they would occur so distant as to make it impracti-
cable ; the following method is preferable for this purpose :
G being the last board obtained by the above method, ex-
tend the curve of its inner edge until it meets the axis, D B,
in e ; from 3, through e, draw 3 f, meeting the arc A B in f ;
join f and 4, f and 5 , and /and 6, cutting the axis, D B, in s, n,
and m ; from 4, 5, and 6 draw lines parallel to A C and cutting
the axis in c, p, and r ; make c 4 (Fig. 101) equal to c 4 in the pre-
vious figure, and c s equal to c s also in the previous figure ;
then describe the inner edge of the board //, according to
Art. 516; the outer edge can be obtained by gauging from
the inner edge. Tn like manner proceed to obtain the next
DESIGNS OF BRIDGES.
223
board — taking/ 5 for half the chord, and p n for the height
of the segment. Should the segment be too large to be de-
scribed easily, reduce it by finding intermediate points in the
curve, as at Art. 515.
241. — Polygonal Dome: Form of Angle-Rib. — To ob-
tain the shape of this rib, let A G H (Fig. 102) be the plan of
a given dome, and C D a vertical section taken at the line
ef. From i, 2, 3, etc., in the arc CD draw ordinates, paral-
lel to A D, to meet f-G ; from the points of intersection on
fG draw ordinates at right angles to/ G\ make s i equal
to o i, s 2 equal to o 2, etc. ; then GfB, obtained in this way,
will be the angle-rib required. The best position for the
sheathing-boards for a dome of this kind is horizontal, but if
they are required to be bent from the base to the vertex,
their shape may be found in a similar manner to that shown
at Fig. 99.
BRIDGES.
FIG. 103
242.— Bridges.— Of plans for the construction of bridges,
perhaps the following are the most useful. Fig. 103 shows a
method of constructing wooden bridges where the banks
of the river are high enough to permit the use of the tie-
beam, a b. The upright pieces, c d, are notched and bolted
on in pairs, for the support of the tie-beam. A bridge ot
this construction exerts no lateral pressure upon the abut-
ments. This method may be employed even where the banks
of the river are low, by letting the timbers for the roadway
rest immediately upon the tie-beam. In this case the irame-
work above will serve the purpose of a railing.
224
CONSTRUCTION.
Fig. 104 exhibits a wooden bridge without a tie-beam.
Where staunch buttresses can be obtained this method may
be recommended ; but if there is any doubt ot their stability,
it should not be attempted, as it is evident that such a sys-
tem of framing is capable of a tremendous lateral thrust.
FIG. 104.
243, — Bridge§: Built-Rib. — Fig. 105 represents a bridge
with a built-rib (see Art. 231) as a chief support. The curve
of equilibrium will not differ much from that of a parabola ;
this, therefore, may be used — especially if the rib is made
FIG. 105.
gradually a little stronger as it approaches the buttresses.
As it is desirable that a bridge be kept low, the following
table is given to show the least rise that may be given to the
rib.
Span in Feet.
Least Rise in Feet
Span in Feet.
Least Rise in Feet
Span in Feet. ! Least Rise in Feet
(
30
o-5
1 2O
7
280
24
40
0-8
I4O
8
3OO
28
50
1-4
1 60
10
320
32
60
2
1 80
ii
350
39
70
21
200
12
380
47
80
3
2 2O
14
400
53
90
4
240
17
100
5
260
20
i
W
((UNIVERSITY
DIMENSIONS OF THE BUILT-Rl]
The rise should never be made less than this, but in all
cases greater if practicable ; as a small rise requires a greater
quantity of timber to make the bridge equally strong. The
greatest uniform weight with which a bridge is likely to be
loaded is, probably, that of a dense crowd of people. This
may be estimated at 70 pounds per square foot, and the fram-
ing and gravelled roadway at 230 pounds more ; which
amounts to 300 pounds on a square foot. The following
rule, based upon this estimate, may be useful in determining
the area of the ribs.
Rule LXVIL— Multiply the width of the bridge by the
square of half the span, both in feet, and divide this pro-
duct by the rise in feet multiplied by the number of ribs ;
the quotient multiplied by the decimal o-oon will give the
area of each rib in feet. When the roadway is only planked,
use the decimal 0-0007 instead of o-ooii,
Example. — What should be the area of the ribs for a
bridge of 200 feet span, to rise 15 feet and be 30 feet wide,
with three curved ribs ? The half of the span is 100, and
its square is 10000 ; this multiplied by 30 gives 300000, and
15 multiplied by 3 gives 45; then 300000 divided by 45
gives 6666|, which multiplied by o-ooii gives 7-333 feet or
1056 inches for the area of each rib. Such a rib may be 24
inches thick by 44 inches deep, and composed of 6 pieces,
2 in width and 3 in depth.
The above rule gives the area of a rib that would be
requisite to support the greatest possible tiniform load.
But in large bridges, a variable load, such as a heavy wagon,
is capable of exerting much greater strains ; in such cases,
therefore, the rib should be made larger.*
In constructing these ribs, if the span be not over 50 feet,
each rib may be made in two or three thicknesses of timber
(three thicknesses is preferable), of convenient lengths bolted
together ; but in larger spans, where the rib will be such as
to render it difficult to procure timber of sufficient breadth,
they may be constructed by bending the pieces to the proper
curve and bolting them together. In this case, where tim-
* See Tredgold's Carpentry by Hurst, Arts. 174 to 177.
226
CONSTRUCTION.
her of sufficient length to span the opening- cannot be ob-
tained, and scarfing is necessary, such joints must be made
as will resist both tension and compression (see Fig. 1 14).
To ascertain the greatest depth for the pieces which compose
the rib, so that the process of bending may not injure their
elasticity, multiply the radius of curvature in feet by the
decimal 0-05, and the product will be the depth in inches.
Example. — Suppose the curve of the rib to be described
with a radius of 100 feet, then what should be the depth ? The
radius in feet, 100, multiplied by 0-05 gives a product of 5
inches. White pine or oak timber 5 inches thick would
freely bend to the above curve ; and if the required depth
of such a rib be 20 inches, it would have to be composed of at
least 4 pieces. Pitch pine is not quite so elastic as white
pine or oak — its thickness may be found by using the deci-
mal 0-046 instead of 0-05.
FIG. 106.
244. — Bridges: Framed Rib. — In spans of over 250
feet, SL framed r\by as in Fig. 106, would be preferable to the
foregoing. Of this, the upper and the lower edges are
formed as just described, by bending the timber to the proper
curve. The pieces that tend to the centre of the curve,
called radials, are notched and bolted on in pairs, and the
cross-braces are halved together in the middle, and abut end
to end between the radials. The distance between the ribs
of a bridge should not exceed about 8 feet. The roadway
should be supported by vertical standards bolted to the ribs
THE ROADWAY AND ABUTMENTS. 22/
at about every 10 to 15 feet. At the place where they rest
on the ribs, a double, horizontal tie should be notched and
bolted on the back of the ribs, and also another on the un-
derside ; and diagonal braces should be framed between the
standards, over the space between the ribs, to prevent lat-
eral motion. The timbers for the roadway may be as light
as their situation will admit, as all useless timber is only an
unnecessary load upon the arch.
245. — Bridges: Roadway. — If a roadway be 18 feet
wide, two carriages can pass without inconvenience. Its
width, therefore, should be either 9, 18, 27, or 36 feet, ac-
cording to the amount of travel. The width of the foot-
path should be two feet for every person. When a stream
of water has a rapid current, as few piers as practicable
should be allowed to obstruct its course ; otherwise the
bridge will be liable to be swept away by freshets. When
the span is not over 300 feet, and the banks of the river are
of sufficient height to admit of it, only one arch should be
employed. The rise of the arch is limited by the form of
the roadway, and by the height of the banks of the river
(see Art. 243). The rise of the roadway should not exceed
one in 24 feet, but as the framing settles about one in 72, the
roadway should be framed to rise one in 18, that it may be
one in 24 after settling. The commencement of the arch at
the abutments — the spring, as it is termed — should not be
below high-water mark ; and the bridge should be placed at
right angles with the course of the current.
246. — Bridges: Abutments. — The best material for the
abutments and piers of a bridge is stone ; and no other
should be used. The following rule is to determine the ex-
tent of the abutments, they being rectangular, and built with
stone weighing 120 pounds to a cubic foot.
Rule LXVIII — Multiply the square of the height of the
abutment by 160, and divide this product by the weight of a
square foot of the arch, and by the rise of the arch ; add
unity to the quotient, and extract the square root. Dimin-
ish the square root by unity, and multiply the root so dimin-
228 CONSTRUCTION.
ished by half the span o-f the arch, and by the weight of a
square foot of the arch. Divide the last product by 120
times the height of the abutment, and the quotient will be
the thickness of the abutment.
Example. — Let the height of the abutment from the base
to the springing of the arch be 20 feet, hall the span 100 feet,
the weight of a square foot of the arch, including the great-
est possible load upon it, 300 pounds, and the rise of the arch
18 feet: what should be its thickness? The square of the
height of the abutment, 400, multiplied by 160 gives 64000,
and 300 by 18 gives 5400; 64000 divided by 5400 gives a
quotient of 11-852; one added to this makes 12-852, the
square root of which is 3-6 ; this, less one is 2-6 ; this mul-
tiplied by 100 gives 260, and this again by 300 gives 78000;
this divided by 120 times the height of the abutment, 2400,
gives 32 feet 6 inches, the thickness required.
The dimensions of a pier will be found by the same rule ;
for, although the thrust of an arch may be balanced by an
adjoining arch when the bridge is finished, and while it re-
mains uninjured, yet, during the erection, and in the event
of one arch being destroyed, the pier should be capable of
sustaining the entire thrust of the other.
Piers are sometimes constructed of timber their princi-
pal strength depending on piles driven into the earth ; but
such piers should never be adopted where it is possible to
avoid them ; for, being alternately wet and dry, they decay
much sooner than the upper parts of the bridge. Spruce
and elm are considered good for piles. Where the height
from the bottom of the river to the roadway is great, it is a
good plan to cut them off at a little below low-water mark,
cap them with a horizontal tie, and upon this erect the posts
for the support of the roadway. This method cuts off the
pan that is continually wet from that which is only occa-
sionally so, and thus affords an opportunity for replacing the
upper part. The pieces which are immersed will last a
great length of time, especially when of elm ; for it is a
well-established fact that timber is less durable when subject
to alternate dryness and moisture than when it is either con-
tinually wet or continually dry. It has been ascertained that
CENTRING FOR BRIDGES.
229
the piles under London Bridge, after having been driven
about 600 years, were not materially decayed. These piles
are chiefly of elm, and wholly immersed.
247. — Centre§ for Stone Bridges. — Fig. 107 is a design
for a centre for a stone bridge where intermediate supports,
as piles driven into the bed of the river, are practicable. Its
timbers are so distributed as to sustain the weight of the
arch-stones as they are being laid, without destroying the
original form of the centre ; and also to prevent its destruc-
tion or settlement, should any of the piles be swept away.
The most usual error in badly-constructed centres is that
the timbers are disposed so as to cause the framing to
rise at the crown during the laying of the arch-stones up
FIG. 107.
the sides. To remedy this evil, some have loaded the crown
with heavy stones ; but a centre properly constructed will
need no such precaution.
Experiments have shown that an arch-stone does not press
upon the centring until its bed is inclined to the horizon at
an angle of from 30 to 45 degrees, according to the hardness
of the stone, and whether it is laid in mortar or not. For
general purposes, the point at which the pressure com-
mences may be considered to be at that joint which forms
an angle of 32 degrees with the horizon. At this point the
pressure is inconsiderable, but gradually increases towards
the crown. The following table gives the portion of the
weight of the arch-stones that presses upon the framing at
the various angles of inclination formed by the bed of the
230 CONSTRUCTION.
stone with the horizon. The pressure perpendicular to the
curve is equal to the weight of the arch-stone multiplied by
the decimal —
•o, when the angle of inclination is 32 degrees.
.04 " " " 34
• 08 " " " 36 "
• 12 " " " 38 "
•17 " " " 40 "
•21 " " " 42 "
-25 " - » 44
• 29 " " " 46 "
• 33 " ;" " 48 "
• 37 " " " 50 "
• 4 " " " 52 "
•44 " 4t " 54 "
•48 " " " 56
• 52 " " u 58 "
-54 " " " 60 "
From this it is seen that at the inclination of 44 degrees the
pressure equals one quarter the weight of the stone ; at 57
degrees, half the weight ; and when a vertical line, as a b
(Fig. 1 08), passing through the centre of
gravity of the arch-stone, does not fall
within its bed, c d, the pressure may be con-
sidered equal to the whole weight of the
stone. This will be the case at about 60
degrees, when the depth of the stone is
double its breadth. The direction of these
pressures is considered in a line with the radius of the curve.
The weight upon a centre being known, the pressure may be
estimated and the timber calculated accordingly. But it
must be remembered that the whole weight is never placed
upon the framing at once — as seems to have been the idea
had in view by the designers of some centres. In building
the arch, it should be commenced at each buttress at the
same time (as is generally the case), and each side should
progress equally towards the crown. In designing the fram-
CENTRE FOR A STONE BRIDGE.
231
ing, the effect produced by each successive layer of stone
should be considered. The pressure of the stones upon one
side should, by the arrangement of the struts, be counter-
poised by that of the stones upon the other side.
Over a river whose stream is rapid, or where it is neces-
sary to preserve an uninterrupted passage for the purposes
of navigation, the centre must be constructed without in-
termediate supports, and without a continued horizontal tie
at the base ; such a centre is shown at Fig. 109. In laying
the stones from the base up to a and c, the pieces bd and
bd act as ties to prevent any rising at b. After this, while
the stones are being laid from a and from c to b, they act as
struts; the piece f g is added for additional security.
Upon this plan, with some variation to suit circumstances,
FIG. 109.
centres may be constructed for any span usual in stone-
bridge building.
In bridge centres, the principal timbers should abut, and
not be intercepted by a suspension or radial piece between.
These should be in halves, notched on each side and bolted.
The timbers should intersect as little as possible, for the
more joints the greater is the settling ; and halving them
together is a bad practice, as it destroys nearly one half the
strength of the timber. Ties should be introduced across,
especially where many timbers 'meet ; and as the centre is
to serve but a temporary purpose, the whole should be de-
signed with a view to employ the timber afterwards for
other uses. For this reason, all unnecessary cutting should
be avoided.
232 CONSTRUCTION.
Centres should be sufficiently strong- to preserve a
staunch and steady form during the whole process of build-
ing ; for any shaking or trembling will have a tendency to
prevent the mortar or cement from setting. For this pur-
pose, also, the centre should be lowered a trifle immedi-
ately after the key-stone is laid, in order that the stones may
take their bearing before the mortar is set ; otherwise the
joints will open on the underside. The trusses, in centring,
are placed at the distance of from 4 to 6 feet apart, accord-
ing to their strength and the weight of the arch. Between
every two trusses diagonal braces should be introduced to
prevent lateral motion.
In order that the centre may be easily lowered, the
frames, or trusses, should be placed upon wedge-formed
sills, as is shown at d (Fig. 109). These are contrived so as
to admit of the settling of the frame by driving the wedge
d with a maul, or, in large centres, with a piece of timber
mounted as a battering-ram. The operation of lowering a
centre should be very slowly performed, in order that the
parts of the arch may take their bearing uniformly. The
wedge pieces, instead of being placed parallel with the
truss, are sometimes made sufficiently long and laid through
the arch, in a direction at right angles to that shown at Fig.
109. This method obviates the necessity of stationing men
beneath the arch during the process of lowering ; and was
originally adopted with success soon after the occurrence of
an accident, in lowering a centre, by which nine men were
killed.
To give some idea of the manner of estimating the pres-
sures, in order to select timber of the proper scantling, cal-
culate the pressure (Art. 247) of the arch-stones from i to b
(Fig. 109), and suppose half this pressure concentrated at a,
and acting in the direction a f. Then, by the parallelogram
of forces (Art. 71), the strain in the several pieces compos-
ing the frame bda may be computed. Again, calculate
the pressure of that portion of the arch included between a
and c, and consider half of it collected at £, and acting in a
vertical direction ; then, by the parallelogram of forces, the
pressure on the beams bd and £</may be found. Add the
JOINTS OF THE ARCH-STONES.
233
pressure of that portion of the arch which is included be-
tween i and b to half the weight of the centre, and consider
this amount concentrated at d, and acting in a vertical direc-
tion ; then, by constructing the parallelogram of forces, the
pressure upon dj may be ascertained.
The strains having been obtained, the dimensions of the
several pieces in the frames £#</and bed may be found by
computation, as directed in the case of roof-trusses, from
Arts. 226 to 229. The tie-beams b d, b d, if made of suffi-
cient size to resist the compressive strain acting upon them
from the load at b, will be more than large enough to resist
the tensile strain upon them during the laying of the first
part of the arch-stones below a and c.
248.— Arcli-Stones : JToint§.— In an arch, the arch-stones
are so shaped that the joints between them are perpendicu-
lar to the curve of the arch, or to its tangent at the point at
which the joint intersects the curve. In a circular arch, the
FIG. no.
joints tend toward the centre of the circle ; in an elliptical
arch, the joints may be found by the following process :
To find the direction of the joints for an elliptical arch ;
FIG. in.
a joint being wanted at a (Fig. no), draw lines from that
point to the foci, /and/; bisect the angle faf with the
line a b ; then a b will be the direction of the joint.
234 CONSTRUCTION.
To find the direction of the joints for a parabolic arch ;
a joint being- wanted at a (Fig. 1 1 1), draw a e at right angles
to the axis eg; make eg equal to c e, and join a and g\
draw a h at right angles to ag\ then a h will be the
direction of the joint. The direction of the joint from b is
found in the same manner. The lines a gaud b f are tan-
gents to the curve at those points respectively ; and any
number of joints in the curve may be obtained by first
ascertaining the tangents, and then drawing lines at right
angles to them. (See Art. 462.)
JOINTS.
24-9. — Timber Joints. — The joint shown in Fig. 112 is
simple and strong ; but the strength consists wholly in the
bolts, and in the friction of the parts produced by screwing
the pieces firmly together. Should the timber shrink to
FIG. ii2.
even a small degree, the strength would depend altogether
on the bolts. It would be made much stronger by indent-
ing the pieces together, as at the upper edge of the tie-beam
FIG. 113.
in Fig. 113, or by placing keys in the joints, as at the lower
edge in the same figure. This process, however, weakens
the beam in proportion to the depth of the indents.
FIG. 114
Fig. 114 shows a method of scarfing, or splicing, a tie-
beam without bolts. The keys are to be of well-seasoned,
hard wood, and, if possible, very cross-grained. The addi-
THE SPLICING OF TIMBER. 235.
tion of bolts would make this a very strong splice, or even
white-oak pins would add materially to its strength.
Fig. 1 1 5 shows about as strong a splice, perhaps, as can
well be made. It is to be recommended for its simplicity ;
as, on account of there being no oblique joints in it, it can
be readily and accurately executed. A complicated joint is
the worst that can be adopted ; still, some have proposed
joints that seem to have little else besides complication to
recommend them.
In proportioning the parts of these scarfs, the depths of
FIG. 115.
all the indents taken together should be equal to one third
of the depth of the beam. In oak, ash or elm, the whole
length of the scarf should be six times the depth, or thick-
ness, of the beam, when there are no bolts ; but, if bolts in-
stead of indents are used, then three times the breadth ; and
when both methods are combined, twice the depth of the
beam. The length of the scarf in pine and similar soft
woods, depending wholly on indents, should be about 12
times the thickness, or depth, of the beam ; when depend-
ing wholly on bolts, 6 times the breadth ; and when both
methods are combined, 4 times the depth.
FIG. 116.
Sometimes beams have to be pieced that are required
to resist cross-strains — such as a girder, or the tie-beam of a
roof when supporting the ceiling. In such beams, the
fibres of the wood in the upper part are compressed ; and
therefore a simple butt joint at that place (as in Fig. 116)
is far preferable to any other. In such case, an oblique
joint is the very worst. The under side of the beam being
in a state of tension, it must be indented or bolted, or both ;
and an iron plate under the heads of the bolts gives a great
addition of strength.
236 CONSTRUCTION.
Scarfing requires accuracy and care, as all the indents
should bear equally ; otherwise, one being strained more
than another, there would be a tendency to splinter off the
parts. Hence the simplest form that will attain the object
is by far the best. In all beams that are compressed end-
wise, abutting joints, formed at right angles to the direction
of their length, are at once the simplest and the best. For a
temporary purpose, Fig. 112 would do very well ; it would
be improved, however, by having a piece bolted on all four
sides. Fig. 113, and indeed each of the others, since they
have no oblique joints, would resist compression well.
In framing one beam into another for bearing purposes,
such as a floor-beam into a trimmer, the best place to make
the mortise in the trimmer is in the neutral line (Arts. 120,
121), which is in the middle of its depth. Some have
thought that, as the fibres of the upper edge are compressed,
FIG. 117.
a mortise might be made there, and the tenon driven in
tight enough to make the parts as capable of resisting the
compression as they would be without it ; and they have
therefore concluded that plan to be the best. This could
not be the case, even if the tenon would not shrink ; for a
joint between two pieces cannot possibly be made to resist
compression so well as a solid piece without joints. The
proper place, therefore, for the mortise is at the middle of
the depth of the beam ; but the best place for the tenon, in
the floor-beam, is at its bottom edge. For the nearer this
is placed to the upper edge, the greater is the liability for it
to splinter off; if the joint is formed, therefore, as at Fig. \ 17,
it will combine all the advantages that can be obtained. Dou-
ble tenons are objectionable, because the piece framed into
is needlessly weakened, and the tenons are seldom so accu-
rately made as to bear equally. For this reason, unless the tusk
THE FRAMING IN A ROOF-TRUSS.
237
at a in the figure fits exactly, so as to bear equally with the
tenon, it had better be omitted. And in sawing the shoulders
care should be taken not to saw into the tenon in the least,
as it would wound the beam in the place least able to bear it.
Thus it will be seen that framing weakens both pieces,
more or less. It should, therefore, be avoided as much as
possible , and where it is practicable one piece should rest
upon the other, rather than be framed into it. This re-
mark applies to the bearing of floor-beams on a girder, to
the purlins and jack-rafters of a roof, etc.
In a framed truss for a roof, bridge, partition, etc., the
joints should be so constructed as to direct the pressures
through the axes of the several pieces, and also to avoid
every tendency of the parts to slide. To attain this object,
FIG. 118.
FIG. 119.
FIG. 120.
the abutting surface on the end of a strut should be at
right angles to the direction of the pressure ; as at the joint
shown in Fig. 1 18 for the foot of a rafter (see Art. 86), in Fig.
1 19 for the head of a rafter, and in Fig. 120 for the foot of a
strut or brace. The joint at Fig. 118 is not cut completely
across the tie-beam, but a narrow lip is left standing in the
middle, and a corresponding indent is made in the rafter, to
prevent the parts from separating sideways. The abutting
surface should be made as large as the attainment of other
necessary objects will admit. The iron strap is added to
prevent the rafter sliding out, should the end of the tie-
beam, by decay or otherwise, splinter off. In making- the
joint shown at Fig. 119, it should be left a little open at a,
so as to bring the parts to a fair bearing at the settling of
the truss, which must necessarily take place from the shrink-
ing of the king-post and other parts. If the joint is made
fair at first, when the truss settles it will cause it to open at
CONSTRUCTION.
the under side of the rafter, thus throwing the whole pres-
sure upon the sharp edge at a. This will cause an indenta-
tion in the king-post, by which the truss will be made to
settle further ; and this pressure not being in the axis of the
rafter, it will be greatly increased, thereby rendering the
rafter liable to split and break.
FIG. 121.
FIG. 122.
FIG. 123.
If the rafters and struts were made to abut end to end,
as in Figs. 121, 122 and 123, and' the king or queen post
notched on in halves -and bolted, the ill effects of shrinking
would be avoided. This method has been practised with
success in some of the most celebrated bridges and roofs in
Europe ; and, were its use adopted in this country, the un-
seemly sight of a hogged ridge would seldom be met with.
FIG. 124.
FIG. 125.
A plate of cast-iron between the abutting surfaces will
equalize the pressure.
Fig. 124 is a proper joint for a collar-beam in a small
roof : the principle shown here should characterize all tie-
joints. The dovetail joint, although extensively practised
in the above and similar cases, is the very worst that can be
employed. The shrinking of the timber, if only to a small
WHITE-OAK PINS AND IRON STRAPS. 239
degree, permits the tie to withdraw — as is shown at Fig.
12$. The dotted line shows the position of the tie after it
has shrunk.
Locust and white-oak pins are great additions to the
strength of a joint. In many cases they would supply the
place of iron bolts ; and, on account of their small cost, they
should be used in preference wherever the strength ef iron
is not requisite. In small framing, good cut nails are of
great service at the joints ; but they should not be trusted
to bear any considerable pressure, as they are apt to be
brittle. Iron straps are seldom necessary, as all the joinings
in carpentry may be made without them. They can be
used to advantage, however, at the foot of suspending-pieces,
and for the rafter at the end of the tie-beam. In roofs for
ordinary purposes, the iron straps for suspending-pieces
may be as follows : When the longest unsupported part of
the tie-beam is —
10 feet, the strap may' be i inch wide by T3T thick.
15 " " " i£ " • " i "
2Q « « « 2 u « JL "
In fastening a strap, its hold on the suspending-piece will be
much increased by turning its ends into the wood. Iron
straps should be protected from rust ; for thin plates of iron
decay very soon, especially when exposed to dampness.
For this purpose, as soon as the strap is made let it be
heated to about a blue heat, and, while it is hot, pour over
its entire surface raw linseed oil, or rub it with beeswax.
Either of these will give it a coating which dampness will
not penetrate.
SECTION III.— STAIRS.
250. — §tair§ : General Requirements. — The STAIRS is
that commodious arrangement of steps in a building by
which access is obtained from one story to another. Their
position, form, and finish, when determined with discrimi-
nating taste, add greatly to the comfort and elegance of a
structure. As regards their position, the first object should
be to have them near the middle of the building, in order
that they may afford an equally easy access to all the rooms
and passages. Next in importance is light ; to obtain which
they would seem to be best situated near an outer wall, in
which windows might be constructed for the purpose ; yet
a skylight, or opening in the roof, would not only provide
light, and so secure a central position for the stairs, but may
be made, also, to assist materially as an ornament to the
building, and, what is of more importance, afford an oppor-
tunity for better ventilation.
All stairs, especially those of the most important build-
ings, should be erected of stone or some equally durable and
fire-resisting material, that the means of egress from a burn-
ing building may not be too rapidly destroyed.
Winding stairs, or those in which the direction is gradu-
ally changed by means of winders, or steps which taper in
width, are interesting by reason of the greater skill required
in their construction ; but are objectionable, for the reason
that children are exposed to accident by their liability to fall
when passing over the narrow ends of the steps. Stairs of
this kind should be tolerated only where there is not suffi-
cient space for those with flyers, or steps of parallel width.
Stairs in one long continuous flight are also objection-
able. Platforms or landings should be introduced at inter-
vals, so that any one flight may not contain more than about
twelve or fifteen steps.
The width of stairs should be in accordance with the im-
KHORSABAD.— ASSYRIAN TEMPLE, RESTORED.
THE GRADE OF STAIRS. 241
portance of the building in which they are placed, varying
from 3 to 12 feet. Where two persons are expected to pass
each other conveniently the least width admissible is 3 feet.
Still, in crowded cities, where land is valuable, the space
allowed for passages is correspondingly small, and in these
stairs are sometimes made as narrow as 2\ feet.
From 3 to 4 feet is a suitable width for a good dwelling ;
while 5 feet will be found ample for stairs in buildings occu-
pied by many people ; and from 8 to 12 feet is sufficient for
the width of stairs in halls of assembly.
To avoid tripping or stumbling, care should be exer-
cised, in the planning of a stairs, to secure an even grade.
To this end, the nosing, or outer edge, of each step should be
exactly in line with all the other nosings. In stairs com-
posed of both flyers and winders, precaution in this regard
is especially needed. In such stairs, the steps — flyers and
winders alike — should be of one width on the line along
which a person would naturally walk when having his hand
upon the rail. This tread-line, consequently, would be paral-
lel with the hand-rail, and is usually taken at a distance of
from 1 8 to 20 inches from the centre of it. In the plan of
the stairs this tread-line should be drawn and divided into
equal parts, each part being the tread, or width of a flyer
from the face of one riser to the face of the next.
251. — The Grade of Stairs. — The extra exertion required
in ascending a staircase over that for walking on level
ground is due to the weight which a person at each step is
required to lift ; that is, the weight of his own body. Hence
the difficulty of ascent will be in proportion to the height of
each step, or to the rise, as it is termed. To facilitate the
operation of going up stairs, therefore, the -risers should be
low. The grade of a stairs, or its angle of ascent, depends
not only upon the height of the riser, but also upon the
width of the step ; and this has a certain relation to the
riser ; for the width of a step should be in proportion to the
smallness of the angle of ascent.
The distance from the top of one riser to the top of the
next is the distance travelled at each step taken, and this dis-
242 STAIRS.
tance should vary as the grade of the stairs ; for a person
who in climbing a ladder, or a nearly vertical stairs, can
travel only 12 inches, or less, at a step, will be able with
equal or greater facility to travel at least twice this distance
on level ground. The distance travelled, therefore, should
be in proportion inversely to the angle of ascent ; or, the di-
mensions of riser and step should be reciprocal : a low rise
should have a wide step, and a high rise a narrow step.
252. — Pitch-Board : Relation of Rise to Tread. — Among
the various devices for determining the relation of the rise
to the tread, or net width of step, one is to make the sum of
the two equal to 18 inches.
For example, for a rise of 6 inches the tread should be
12, for 7 inches the tread should be n ; or —
6 + 12 = 18 8 + 10 = 18
6J + iii= 18 8i + 9i = 18
7 + ii = 18 9 + 9 = 18
;£+ ioi= 18 Qj + 8i= 18
This rule is simple, but the results in extreme cases are not
satisfactory. If the ascent of a stairs be gradual and easy,
the length from the top of one rise to that of another, or the
hypothenuse of the pitch-board, may be proportionally long ;
but if the stairs be steep, the length must be shorter.
There is a French method, introduced by Blondel in his
Cours d' Architecture. It is referred to in Gwilt's Encyclo-
pedia, Art. 2813. •
This method is based upon the assumed distance of 24
inches as being a convenient step upon level ground, and
upon 12 inches as the most convenient height to rise when
the ascent is vertical. These are French inches, old system.
The 24 inches French equals about 25^ inches English.
With these distances as base and perpendicular, a right-
angled triangle is formed, which is used as a scale upon
which the proportions of a pitch-board are found. For
example, let a line be drawn from any point in the hypothe-
nuse of this triangle to the right angle of the triangle ; then
this line will equal the length of the pitch-board, along the
PROPORTIONS OF THE PITCH-BOARD. 243
rake, for a stairs having a grade equal to the angle formed
by this line and the base-line of the scale.
In the absence of the triangular scale, the lengths of the
pitch-boards, as found by this rule, may be computed by this
expression—
W=2$3r-2A; (107.)
in which W equals the tread, or base of the pitch-board, and
h the riser, or its perpendicular height.
For example, let // = 6 ; then —
^=25^-2x6=13^.
This result is greater than would be proper in some cases.
The length of the hypothenuse of the pitch-board should
be proportional not only to the angle of ascent (Art. 251), but
also to the strength and height of the class of people who
are to use the stairs. Tall and strong persons will take
longer steps than short and feeble people. The hypothe-
nuse of the pitch-board should be made in proportion to
the distance taken at a step on level ground by the persons
who are to use the stairs.
If people are divided into two classes, one composed of
robust workmen and the other of delicate women and in-
firm men, then there may be two scales formed for the pitch-
boards of stairs — one to be used for shops and factories, and
the other for dwellings. The distance on level ground trav-
elled per step, by men, varies from about 26 to 32 inches, or
on an average 28 inches. The height to which men are
accustomed to rise on ladders is from 12 to 16 inches at each
step, or on the average 14 inches.
With these dimensions, therefore, of 14 and 28 inches, a
scale may be formed for pitch-boards for stairs, in buildings
to be used exclusively by robust workmen. And with 12
and 24 inches another scale may be formed for pitch-boards
for stairs, in buildings to be used by women and feeble
people. These two scales are both shown in Fig. 126.
They are made thus : Let C A B be a right angle. Make A
B equal to 28 inches, and A C equal to 14 ; then join B and
244
STAIRS.
C. At right angles to C B, from A, draw A F\ then with
A F for radius describe the arc F G. Then a' line, as A K
or A L, drawn from A at any angle with A B and limited by
the line G FB will give the length of the hypothenuse of
the pitch-board, for shop stairs of a grade equal to the angle
which said line makes with A B. From K, perpendicular to
A B, draw K N "; then K N will be the proper riser for a
pitch-board of which A N is the tread. So, likewise, L M
will be the appropriate riser for the tread A M. The arc F G
is introduced to limit the rake-line of pitch-boards occur-
ring between F and C, in order to avoid making them longer
than the one at F. The scale for the stairs for dwellings is
made in the same manner ; A D — 24 inches being the base,
A E = 12 inches the rise, and J H D the line limiting the
rake-lines of pitch-boards.
M
N
FIG. 126.
To compute the length of risers and treads, we have for
the scale for shops, for those occurring between F and B —
r = 4(28-/): (108.)
fi=28-2rj (109.)
and for those between F and G, we have —
(108, A.)
(109, A.)
r=
-/2;
1/756^8-
For the scale for dwellings, we have, for those occurring
between H and D —
r =i(24-/); (108, B.)
t = 24 — 2 r ; (109, B.)
STAIRS FOR SHOPS AND FOR DWELLINGS.
and for those between H and J, we have —
t =
245
, c.)
(109, C.)
where, in each equation, r represents the riser, and t the
tread, or net step.
By these formulae, the following tables have been com-
puted :
STAIRS FOR SHOPS.
Rise.
Tread.
Ratio— Rise to Tread.
Rise.
Tread.
Ratio— Rise to Tread.
2-
24-
to 12-
7-40
13-20
to -78
3'
22-
" 7-33
7-60
12-80
" -68
3-50
21-
" 6-
7-80
12-40
" -59
4-
20-
" 5-
8-
12-
•50
4-50
19.
" 4-22
8-20
II-6
" -41
5'
18-
" 3'6o
8-50
II-
•29
5-4
17-20
3-19
8-80
10-40
" -18
5-7
16-60
' 2-91
9'
10-
" -ii
6-
16-
' 2-67
9-30
9-40
" -01
6-25
I5-50
: 2-48
9-60
8-80
" 0-92
6-50
15-
' 2-31
10-
8-
. ' 0-80
6-70
14-60
' 2-18
10-50
7'
4 0-67
6-90
14-20
' 2-06
ii-
6-
1 °'55
7-
14-
' 2-
11-50
4-95
' 0-43
7-20
13-60
' I-89
12-
3-58
i ' 0-30
STAIRS FOR DWELLINGS.
Rise.
Tread.
Ratio— Rise to Tread.
Rise.
Tread.
Ratio — Rise to Tread.
2-
2O-
I to 10-
7-40
9-20
i .to -24
3"
18-
I " 6-
7-50
9'
I " -2O
3-50
17-
i ' 4-86
7'60
8-80
" .16
4'
16-
I ' 4-
7-70
8-60
" -12
4-50
15-
' 3-33
7-80
8-40
" -08
5'
14-
2-80
7-90
8-20
" -04
5-40
13-20
' 2-44
8-
8-
•
5-70
12-60
" 2-21
8-10
7-80
" 0-96
6-
12-
" 2-
8-30
7-40
" 0-89
6-25
II-5O
" -84
8-50
7-
" 0-82
6-50
II-
" -69
8-75
6-50
" o-74
6-75
IO-5O
' -56
9'
6-
" 0-67
7-
10-
•43
9-30
5-40
" 0-58
7-10
9-80
' -38
9-60
4-80
" 0-50
7-20
9-60
•33
10-
3.90
" 0-39
7-30
9.40
-29
10-50
2 -2O
' 0-21
These tables will be useful in determining questions in-
246 STAIRS.
volving the proportion between the rise and tread of a
pitch-board.
For stairs in which the run is limited, to determine the
number of risers which would give an easy ascent : Divide
the run by the height, and find in the proper table, above,
the ratio nearest to the quotient, and in a line with this ratio,
in the second column to the left, will be found the corre-
sponding riser. With this divide the rise in inches ; the quo-
tient, or the nearest whole number thereto, will be the required
number of risers in the stairs.
Example. — For the stairs in a dwelling, let the rise be 12'
8", or I524nches. Let the run between the extreme risers
be 17' 2". To this, for the purpose of obtaining the correct
angle of ascent, by having an equal number of risers and
treads, add, for one more tread, say 10 inches, its probable
width; thus making the total run 1 8 feet, or 216 inches.
Thus we have for the run 216, and for the rise 152. Divid-
ing the former by the latter gives i -42 nearly. In the table
of stairs for dwellings, the ratio nearest to this is I -43, and in
the line to the left, in the second column, is 7, the approxi-
mate size of riser appropriate to this case. Dividing the
rise, 152 inches, by this 7, we have 2 if as the quotient.
This is nearer to 22 than to 21 ; therefore, the number of
risers required is 22.
When the number of risers is determined, then the rise
divided by this number will give the height of each riser ;
thus, in the above case, the rise is 152 inches. This divided
by 22 gives 6-909 inches for the height of the riser.
When the height of the riser is known, then, if the run is
unlimited, the width of tread will be found in the proper table
above. For example, if the riser is 7 inches or nearly that,
then in the table of stairs for dwellings, in the next column
to the right, and opposite 7 in the column of risers, is found
10, the approximate width of tread. By the use of equation
(109, B.), the width may be had exactly according to the
scale. For example, equation (109, B.) with 6-91 for the
riser, becomes—
t — 24 — 2 x 6-91 = io- 18,
or about ioT3^ inches.
TO CONSTRUCT THE PITCH-BOARD. 247
When the run is limited and the number of risers is
known, then the width of tread is obtained by dividing the
run by the number of treads. There are always of treads
one less than there are of risers, in each flight.
253. — Dimensions of the Pitch-Board. — The first thing
in commencing to build a stairs is to make the //fc/j-board ;
this is done in the following manner : Obtain very accurate-
ly, in feet and inches, the rise, or perpendicular height, of the
story in which the stairs are to be placed. This must be
taken from the top of the lower floor to the top of the upper
floor. Then, to obtain the number of rises and treads and
their size, proceed as directed in Art. 252. Having obtained
these, the pitch-board may be made in the following man-
ner: Upon a piece of well-seasoned board about -| of an
inch thick, having one edge jointed
straight and square, lay the corner
of a steel square, as shown at Fig. 127.
Make a b equal to the riser, and b c
equal to the tread ; mark along the
edges with a knife, and cut by the FlG< I27'
marks, making the edges of the pitch - board perfectly
square. The grain of the wood should run in the direction
indicated in the figure, because, in case of shrinkage, the
rise and the tread will be equally affected by it. When a
pitch-board is first made, the dimensions of the riser and
tread should be preserved in figures, in order that, in case
of shrinkage or damage otherwise, a second may be
made.
254. — The String of a Stairs. — The space required for
timber and plastering under the steps is about 5 inches for
ordinary stairs, or 6 inches if furred ; set a gauge, there-
fore, at 5 or 6 inches, as the case requires, and run it on the
lower edge of the plank, as ab (Fig. 128). Commencing at
one end, lay the longest side of the pitch-board against the
gauge-mark, a b, as at c, and draw by the edges the lines for
the first rise and tread ; then place it successively as at d, e,
248
STAIRS.
and f, until the required number of risers shall be laid down.
To insure accuracy, it is well to ascertain the theoretical
raking length of the pitch-board by computation, as in note
to Art. 536, by getting the square root of the sum of the
squares of the rise and run, and using this by which to
divide the line ab into equal parts.
f
FIG. 128.
255. — Step and Riser Connection. — Fig. 129 represents
a section of step and riser, joined after the most approved
method. In this, a represents the end of a block about 2
FIG. 129,
inches long, two or three of which, in the length of the
step, are glued in the corner. The cove at b is planed up
square, glued in, and stuck or moulded after the glue is
set.
PLATFORM STAIRS.
256. — Platform Stair* : the Cylinder. — A platform stairs
ascends from one story to another in two or more flights,
having platforms or landings between for resting and to
change their direction. This kind of stairs, being simple, is
CYLINDER OF PLATFORM STAIRS.
249
easily constructed, and at the same time is to be preferred
to those with winders, for the convenience it affords in use
(Art. 250). The cylinder may be of
any diameter desirable, from a few
inches to 3 or more feet, but it is
generally small, about 6 inches. It
may be worked out of one solid
piece, but a better way is to glue
together 3 pieces, as in Fig. 130 ; in
which the pieces a, b, and c compose
the cylinder, and d and e represent
parts of the strings. The strings,
after being glued to the cylinder, are secured with screws.
The joining at o and o is the most proper for that kind of
joint.
FIG. 130.
FIG. 131.
257. — Form of Lower Edge of Cylinder. — Find the
stretch-out, de (Fig. 131), of the face of the cylinder, a be,
2?O STAIRS.
according to Art. 524; from d and e draw df and eg at
right angles to d^; draw kg parallel to de, and make hf
and £-2 each equal to one riser; from i and f draw ij and
/£ parallel to hg\ place the tread of the pitch-board at these
last lines, and draw by the lower edge the lines kh and il ;
parallel to these draw m n and op, at the requisite distance
for the dimensions of the string ; from s, the centre of the
plan, draw sq parallel to df\ divide // q and qg each into two
equal parts, as at v and w ; from v and w draw v n and w o
parallel to fd\ join n and o, cutting qs in r ; then the angles
unr and rot, being eased off according to Art. 521, will give
the proper curve for the bottom edge of the cylinder. A
centre may be found upon which to describe these curves,
thus : from u draw u x at right angles to m n ; from r draw
r x at right angles to no ; then x will be the centre for the
curve ur. The centre for the curve rt may be found in a
similar manner. Centres from which to strike these curves
are usually quite unnecessary ; an experienced workman
will readily form the curves guided alone by his practised
eye.
FIG. 132.
258. — Position of the Balu§ters. — Place the centre of
the first baluster, b (Fig. 132), half its diameter from the face
of the riser, cd, and one third its diameter from the end of
the step, e d\ and place the centre of the other baluster, a,
half the tread from the centre of the first. A line through
the centre of the rail will occur vertically over the centres
of the balusters. The usual length of the balusters is 2 feet
5 inches and 2 feet 9 inches respectively, for the short and
long balusters. Their length may be greater than is here
indicated, but, for safety, should never be less. The differ-
ence in length between the short and long balusters is
equal to one half the height of a riser.
CONSTRUCTION OF WINDING STAIRS.
251
259.— Winding stairs: have the steps narrower atone
end than at the other. In some stairs there are steps of
parallel width incorporated with the tapering steps ; in this
case the former are called flyers, and the latter winder:.
260.— Regular Winding Stairs — In Fig. 133, abed rep-
resents the inner surface of the wall enclosing the space
allotted to the stairs, a e the length of the steps, and efgk
the cylinder, or face of the front-string. The line a e is given
as the face of the first riser, and the point / for the limit of
FIG. 133
the last. Make e i equal to 18 inches, and upon o, with o i
for radius, describe the arc i j \ obtain the number of risers
and of treads required to ascend to the floor at j, according
to Art. 252, and divide the arc ij into the same number of
equal parts as there are to be treads : through the points of
division, 1,2, 3, etc., and from the wall-string to the front-
string, draw lines tending to the centre, o : then these lines
will represent the face of each riser, and determine the form
and width of the steps. Allow the necessary projection for
the nosing beyond a e, which should be equal to the thick-
STAIRS.
ness of the step, and then a e I k will be the dimensions for
each step. Make a pitch-board for the wall-string having a k
for the tread, and the rise as previously ascertained : with
this lay out on a thicknessed plank the several risers and
treads, as at Fig. 128, gauging from the upper edge of the
string for the line at which to set the pitch-board.
Upon the back of the string, with a ij-inch dado plane,
make a succession of grooves ij inches apart, and parallel
with the lines for the risers on the face. These grooves
must be cut along the whole length of the plank, and deep
enough to admit of the plank's bending around the curve
abed. Then construct a drum, or cylinder, of any com-
mon kind of stuff, made to fit a curve with a radius the
thickness of the string less than oa ; upon this the string must
be bent, and the grooves filled with strips of wood, called
keys, which must be very nicely fitted and glued in. After
it has dried, a board thin enough to bend around on the out-
side of the string must be glued on from one end to the
other, and nailed with clout-nails. In doing this, be careful
not to nail into any place opposite to where a riser or step is
to enter on the face.
After the string has been on the drum a sufficient time
for the glue to set, take it off, and cut the mortices for the
steps and risers on the face at the lines previously made ;
which may be done by boring with a centre-bit half through
the string, and nicely chiselling to the line. The drum need
not be made to extend over the whole space occupied by the
stairs, but merely so far as requisite to receive one piece of
the wall-string at a time ; for it is evident that more than
One will be required. The front-string may be constructed
in the same manner ; taking e I instead of a k for the tread of
the pitch-board, dadoing it with a smaller dado plane, and
bending it on a drum of the proper size.
261. Winding Stairs : Shape and Position of Timbers. —
The dotted lines in Fig. 133 show the position of the timbers
as regards the plan ; the shape of each is obtained as follows:
In Fig. 134, the line i a is equal to a riser, less the thickness
ot the floor, and the lines 2 m, 3 n, 4 <?, 5 /, and 6 q are each
TIMBERS FOR WINDING STAIRS. 253
equal to one riser. The line a 2 is equal to a m in Fig. 133,
the line m 3 to m n in that figure, etc. In drawing this
figure, commence at a, and make the lines a i and a 2 of the
length above s*pecified, and draw them at right angles to
each other ; draw 2 m at right angles to a 2, and m 3 at
right angles to ;;/ 2, and make 2 m and m 3 of the lengths
as above specified ; and so proceed to the end. Then
through the points i, 2, 3, 4, 5, and 6 trace the line \b\ upon
the points i, 2, 3, 4, etc., with the size of the timber for
radius, describe arcs as shown in the figure, and by these
the lower line may be traced parallel to the upper. This
will give the proper shape for the timber, a b, in Fig. 133 ;
and that of the others may be found in a similar manner. In
ordinary cases, the shape of one face of the timber will be
sufficient, for a good workman can easily hew it to its
proper level by that ; but where great accuracy is desirable,
a pattern for the other side may be found in the same man-
FIG. 134.
ner as for the first. In many cases, the timbers beneath cir-
cular stairs are put up after the stairs are erected, and with-
out previously giving them the required form ; the work-
man in shaping them being guided by the form marked out
by the lower edge of the risers.
262. — Winding Stairs with Flyers : Grade of Front-
String. — In stairs of this kind, if the winders are confined to
the quarter circle, the transition from the winders to the
flyers is too abrupt for convenience, as well as in appear-
ance. To remove this unsightly bend in the rail and string,
it is usual to take in among the winders one or more of the
flyers, and thus graduate the width of the winders to that of
the flyers. But this is not always done so as to secure the
best results. By the method now to be shown, both rail and
strings will be gracefully graded. In Fig. 135, a b repre-
sents the line of the facia along the floor of the upper story,
254 STAIRS.
bee the face of the cylinder, and c d the face of the front-
string. Make gb equal to £ of the diameter of the baluster,
and parallel to a b, b e c, and c d draw the centre-line of the
rail, fg> g k z, and ij\ make gk and gl each equal to half the
width of the rail, and through k and /, parallel to the centre-
line, draw lines for the convex and the concave sides of the
rail ; tangical to the convex side of the rail, and parallel to
k m, draw ;/ o ; obtain the stretch-out, q r, of the semicircle,
k p m, according to Art. 524; extend a b to /, and k m to s;
make c s equal to the length of the steps, and i u equal to 18
inches, and parallel to m p describe the arcs s t and u 6 ;
from / draw / w, tending to the centre of the cylinder ; from
6, and on the line 6 u x, run off the regular tread, as at 5, 4,
3, 2, i, and v\ make ' u x equal to half the arc u 6, and make
the point of division nearest to x, as v, the limit of the par-
allel steps, or flyers ; make r o equal to m z; from o draw o
a** at right angles to ;/ o, and equal to one riser; from a2
draw a* s parallel to n o, and equal to one tread; from s,
through o, draw s b*.
Then from w draw w c2 at right angles to ;/ o, and set up
on the line w c^ the same number of risers that the floor, A,
is above the first winder, B, as at i, 2, 3, 4, 5, and 6 ; through
5 (on the arc 6 u) draw d* e*, tending to the centre of the
cylinder; from e*~ draw erf* at right angles to 110, and
through 5 (on the line w c*} draw g* f2 parallel to n o\
through 6 (on the line zu c2) and/2 draw the line h* b*\
make 6 c~ equal to half a riser, and from c* and 6 draw c'2 i'2
and 6/2 parallel to n o\ make h* i* equal to 7/2/2; from i-
draw i* k* at right angles to i* h-, and from f~ draw /2 k*
at right angles to /2 /*2; upon k~, with £2/2 for radius, de-
scribe the arc /2 *2; make b~ /2 equal to £2/2, and ease oft
the angle at b* by the curve/2 /2. In the figure, the curve
is described from a centre, but as this might be imprac-
ticable in a full-size plan, the curve may be obtained accord-
* In the references a2, />', etc., a new form is introduced for the first time.
During the time taken to refer to the figure, the memory of the form of these
may pass from the mind, while that of the sound alone remains; they may
then be mistaken for a 2, b 2, etc. This can be avoided in reading by giving
them a sound corresponding to their meaning, which is a second, b second, etc.
MOULDS FOR QUARTER-CIRCLE STAIRS.
255
ing to Art. 521. Then from i, 2, 3, and 4 (on the line w c~)
draw lines parallel to n o, meeting the curve in w2, n~, o2,
and /2; from these points draw lines at right angles to ;/ o,
and meeting it in jr2, r2, sz, and /2; from x~ and r2 draw
FIG. 135.
lines tending to z/2, and meeting the convex side of the rail
in j/2 and £2 ; make m ?>2 equal to r s~, and m w* equal to
rt*-, fromj/ W2, and w2, through 4, 3, 2, and i, draw lines
meeting the line of the wall-string in a3, £3, c\ and rt78 ; from
256 STAIRS.
e3, where the centre-line of the rail crosses the line of the
floor, draw e3/3 at right angles to n o, and from/3, through
6, draw f*g~\ then the heavy lines f*g\ e* d\ y* a*, Z* b\
v*c3, w2 d3, and z y will be the lines for the risers, which,
being extended to the line of the front-string, b e c d, will
give the dimensions of the winders and the grading of the
front-string, as was required.
HAND-RAILING.
263.— Hand-Railing for Stairs.— A piece of hand-rail-
' ing intended for the curved part of a stairs, when properly
shaped, has a twisted form, deviating widely from plane sur-
faces. If laid upon a table it may easily be rocked to and
fro, and can be made to coincide with the surface of the
table in only three points. And yet it is usual to cut such
twisted pieces from ordinary parallel-faced plank ; and to
cut the plank in form according to a face-mould, previously
formed from given dimensions obtained from the plan of the
stairs. The shape of the finished wreath differs so widely
from the piece when first cut from the plank as to make it
appear to a novice a matter of exceeding difficulty, if not an
impossibility, to design a face-mould which shall cover accu-
rately the form of the completed wreath. But he will find,
as he progresses in a study of the subject, that it is not only
a possibility, but that the science has been reduced to such
a system that all necessary moulds may be obtained with
great facility. To attain to this proficiency, however, re-
quires close attention and continued persistent study, yet no
more than this important science deserves. The young car-
penter may entertain a less worthy ambition than that of
desiring to be able to form from planks of black-walnut or
mahogany those pieces of hand-railing which, when secured
together with rail-screws, shall, on applying them over the
stairs for which they are intended, be found to fit their
places exactly, and to form graceful curves at the cylinders.
That railing which requires to be placed upon the stairs
before cutting the joints, or which requires the curves or
butt-joints to be refitted after leaving the shop, is discredit-
PRINCIPLES OF HAND-RAILING. 257
able to the workman who makes it. No true mechanic will
be content until he shall be proved able to form the curves
and cut the joints in the shop, and so accurately that no altera-
tion shall be needed when the railing is brought to its place
on the stairs. The science of hand-railing requires some
knowledge of descriptive geometry— that branch of geometry
which has for its object the solution of problems involving
three dimensions by means of intersecting planes. The
method of obtaining the lengths and bevils of hip and valley
rafters, etc., as in Art. 233, is a practical example of descrip-
tive geometry. The lines and angles to be developed in
problems of hand-railing are to be obtained by methods
dependent upon like principles.
264.— Hand-Railing: Definitions; Planes and Solids.
—Preliminary to an exposition of the method for drawing
the face-moulds of a hand-rail wreath, certain terms used in
descriptive geometry need to be denned. Among the tools
used by a carpenter are those well-known implements called
planes, such as the jack-plane, fore-plane, smoothing-plane,
etc. These enable the workman to straighten and smooth
the faces of boards and plank, and to dress them out of
wind, or so that their surfaces shall be true and unwinding.
The term plane, as used in descriptive geometry, however,
refers not to the implement aforesaid, but to the unwinding
surface formed by these implements. A plane in geometry
is defined to be such a surface that if any two points in it be
joined by a straight line, this line will be in contact with the
surface at every point in its length. With like results lines
may be drawn in all possible directions upon such a sur-
face. This can be done only upon an unwinding surface ;
therefore, a plane is an unwinding surface. Planes are
understood to be unlimited in their extent, and to pass freely
through other planes encountered.
The science of stair-building has to do with prisms and
cylinders, examples of which are shown in Figs. 136, 137, and
138. A right prism (Figs. 136 and 137) is a solid standing
upon a horizontal plane, and with faces each of which is a
plane. Two of these faces— top and bottom— are horizontal
258
STAIRS.
and are equal polygons, having their corresponding sides
parallel.
The other faces of the prism are parallelograms, each of
which -is a vertical plane. When the vertical sides of a
prism are of equal width, and in number increased indefi-
nitely, the two polygonal faces of the prism do not differ
essentially from circles, and thence the prism becomes a
cylinder. Thus a right cylinder may be defined to be a
prism, with circles for the horizontal faces (Fig. 138).
FIG. 136.
FIG. 137-
FIG. 138.
265. — Hand - Railing : Preliminary Considerations. —
If within the well-hole, or stair-opening, of a circular stairs a
solid cylinder be constructed of such diameter as shall fill the
well-hole completely, touching the hand-railing at all points,
and then if the top of this cylinder be cut off on a line with
the top of the hand-railing, the upper end of the cylinder
would present a winding surface. But if, instead of cutting
the cylinder as suggested, it be cut by several planes, each
of which shall extend so as to cover only one of the wreaths
of the railing, and be so inclined as to touch its top in three
points, then the form of each of these planes, at its intersec-
tion with the vertical sides of the cylinder, would present
the shape of the concave edge of the face-mould for that
particular piece of hand - railing covered by the plane.
Again, if a hollow cylinder be constructed so as to be in
contact with the outer edge of the hand-railing throughout
its length, and this cylinder be also cut by the aforesaid
FACE-MOULDS FOR HAND-RAILS. 259
planes, then each of said planes at its intersection with this
latter cylinder would present the form of the convex edge
of the said face-mould. A plank of proper thickness may
now have marked upon it the shape of this face-mould, and
the piece covered by the face-mould, when cut from the
plank, will evidently contain a wreath like that over which
the face-mould was formed, and which, by cutting away the
surplus material above and below, may be gradually wrought
into the graceful form of the required wreath.
By the considerations here presented some general idea
may be had of the method pursued, by which the form of a
face-mould for hand-railing is obtained. A little reflection
upon what has been advanced will show that the problem
to be solved is to pass a plane obliquely through a cylinder
at certain given points, and find its shape at its intersection
with the vertical surface of \he cylinder. Peter Nicholson
was the first to show how this might be done, and for the
invention was rewarded, by a scientific society of London,
with a gold medal. Other writers have suggested some
slight improvements on Nicholson's methods. The method
to which preference is now given, for its simplicity ot work-
ing and certainty of results, is that which deals with the
tangents to the curves, instead of with the curves themselves;
so we do not pass a plane through a cylinder, but through a
prism the vertical sides of which are tangent to the cylinder,
and contain the controlling tangents of the face-moulds. The
task, therefore, is confined principally to finding the tangents
upon the face-mould. This accomplished, the rest is easy, as
will be seen.
The method by which is found the form of the top of a
prism cut by an oblique plane will now be shown.
266.— A Pri§m Cut toy an Oblique Plane — A prism is
shown in perspective at Fig. 139, cut by an oblique plane.
The points abed are the angles of the horizontal base, and
abg, bcf, cdcf, and adeg are the vertical sides; while
efbg is the top, the form of which is to be shown.
267.— Form of Top of Pri§m — In Fig. 139 the form of
the top of the prism is shown as it appears in perspective..
26o
STAIRS.
not in its real shape ; this is now to be developed. In Fig.
140, let the sauare a b c d represent by scale the actual form
FIG. 139.
and size of the base, a b cd, of the prism shown in Fig. 1 39.
Make c c, and ddt respectively equal to the actual heights at
FIG. 140.
cf and de, Fig. 139 ; the lines ddt and c c, being set up per-
pendicular to the line dc. Extend the lines dc and dtct until
ILLUSTRATION BY PLANES. 26 1
they meet in h ; join b and h. Now this line b h is the inter-
section of two planes : one, the base, or horizontal plane upon
which the prism stands ; the other, the cutting plane, or the
plane which, passing- obliquely through the prism, cuts it so
as to produce, by intersecting the vertical sides of the prism,
the form b fe g, Fig. 139.
To show that b k is the line of intersection of these two
planes, let the paper on which the triangle dhdt is drawn
(designated by the letter B) be lifted by the point dt and
revolved on the line dk until dt stands vertically over d, and
ct over c\ then B will be a plane standing on the line dh,
vertical to the base-plane A. The point h being in the line
cd extended, and the line cd being in the base-plane A, there-
fore h is in the base-plane A. Now the line dtct represents
the line cf of Fig. 139, and is therefore in the cutting plane ;
consequently the point //, being also in the line dt c, ex-
tended, is also in the cutting plane. By reference to Fig.
139 it will be seen that the point b is in both the cutting and
base planes ; we must therefore conclude that, since the two
points b and h are in both the cutting and base planes, a line
joining these two points must be the intersection of these two
planes. The determination of the line of intersection of the
base and cutting planes is very important, as it is a control-
ling line ; as will be seen in denning the lines upon which
the form of the face-mould depends. Care should therefore
be taken that the method of obtaining it be clearly under-
stood.
It will be observed that the intersecting line bh, being in
the horizontal plane A, is therefore a horizontal line. Also,
that this horizontal line b h being a line in the cutting plane,
therefore all lines upon the cutting plane which are drawn
parallel to b h must also be horizontal lines. The import-
ance of this will shortly be seen. Through a, perpendicular
to bh, draw the line bnd^ and parallel with this line draw
ddini ; on d as centre describe the arc dt dltil ; draw dltll dv
parallel with ddtlJ and extend the latter to dtll ; on d{l as
centre describe the arc dv dnl ; join bn and dul. We now
have three vertical planes which are to be brought into
position around the base-plane A, 'as follows: Revolve B
262 STAIRS.
upon dh, E upon ddit, and C upon bn dtn each until it stands
perpendicular to the plane -A. Then the points dt and dillt
will coincide and be vertically over d\ the points dllt and dv
will coincide and stand vertically over dn ; and ct will cover c.
These vertical planes will enclose a wedge-shaped figure,
lying with one face, b^d^dh, horizontal and coincident with
the base-plane A, and three vertical faces, blt du dti/y ddn d^ djiit,
and hddt. By drawing the figure upon a piece of stout
paper, cutting it out at the outer edges, making creases in
the lines hd, ddtl, dubt/J then folding the three planes B, E,
and C at right angles to A, the relation of the lines will be
readily seen. Now, to obtain the form of the top or cover to
the wedge-shaped figure, perpendicular to bn diti draw b,,h,
and dtlle\ on btl as centre describe the arc hhi ; make dllte
equal to dlt d\ join e and ht. Now the form of the top of the
wedge-shaped figure is shown within the bounds din bi{ ht c.
By revolving this plane D on the line bn dllt until it is at
a right angle to the plane C, and this while the latter is
supposed to be vertical to the plane A, it will be perceived
that this movement will place the plane D on top of the
wedge-shaped figure, and in such a manner as that the point
e will coincide with dlllt d{, and the point ht will fall upon and
be coincident with the point h, and the lines of the cover
will coincide with the corresponding lines of the top edges
of the sides of the figure; for example, the line blidlll is
common to the top and the side C; the line dltle equals dtl d,
which equals dvdttil\ therefore, the line ditte will coincide
with dvdi/u of the side E\ the line eht will coincide with d,h
of the side B\ and the line bl,hl will coincide with the line
btlh. Thus the figure D bounded by blldlllchl will exactly
fit as a cover to the wedge-shaped figure. Upon this cover
we may now develop the form of the top of the prism.
Preliminary thereto, however, it will be observed, as was
before remarked, that lines upon the cutting plane which
are parallel to the intersecting line btl ht are horizontal ;
and each, therefore, must be of the same length as the line
in the base-plane A vertically beneath it. For example, the
line dllt e{ is a line in the cutting plane D, parallel with the
line bltht in the same plane, and this line blllil will (when the
EXPLANATION OF THE DIAGRAMS. 263
cutting plane D is revolved into its proper position) be co-
incident with the intersecting line blt h ; therefore, the line
dtile is a line in the cutting plane D, drawn parallel with the
intersecting line bu h. Now this line dllt e, when in position,
will be coincident with the line dltlld^ which lies vertically
over the line dt,d-ol the base-plane A ; its length, therefore,
is equal to that of the latter. In like manner it may be
shown that the length of any line on the plane D parallel
to btl hn is equal in length to the corresponding line upon
the plane A vertically beneath it.
Therefore, to obtain the form of the top of the prism, we
proceed as follows : Perpendicular to btl dv draw c clu and
aattl\ perpendicular to btldni draw ctllf and equal toc,,c;
on blt as centre describe the arc b bt ; join bt a,n, btf, and al4l e.
Now we have here in plane D the form of the top of the
prism, as shown in the figure bounded by the lines a^'fije.
This will be readily seen when the plane D is revolved into
position. Then the point atll will be vertically over a ; the
point e coincident with dt dltll and vertically over d; the point
/ coincident with c/ and vertically over c ; while bt will coin-
cide with b of the base-plane A.
The figure ani btfe, therefore, represents correctly both
in form and size the top of the prism as it is shown in per-
spective at bfeg, Fig. 139. The line ef, Fig. 140, is equal to
the line dt ct, and so of the other lines bounding the edges of
the figure.
The cutting plane b f e g, Fig. 139, may be taken to repre-
sent the surface of the plank from which the wreath of hand-
railing is to be cut ; the wreath curving around from b to ct
as shown in Fig. 141, the lines b g and ge being tangent to
the curve in the cutting plane; while ab and ad are tan-
gents to the curve on the base plane, or plane of the cylin-
der. The location of the cutting plane, however, is usually
not at the upper surface of the plank, but midway between
the upper and under surfaces. The tangents in the plane
are found to be more conveniently located here for deter-
mining the position of the butt-joints. For a moulded rail
two curved lines, each with a pair of tangents, are required
upon the cutting plane, one for the outer edge of the rail,
264 STAIRS.
and the other for the inner edge ; but for a round rail only
one curve with its tangents is required, as that from b to e
in Fig. 141, which is taken to represent the curved line run-
ning through the centre of the cross-section of the rail. As
an easy application of the principles regarding the prism,
just developed, an example will now be given.
268. — Face-Mould for Hand-Railing of Platform Stairs.
— Let/£ and / ;;/, Fig. 142, represent the central or axial lines
of the hand-rails of the two flights, one above, the other be-
low the platform ; and let the semicircle/df/ be the central
line of the rail around the cylinder at the platform, the risers
at the platform being located at j and /. Vertically over the
platform risers draw ggt ; make grt equal to a riser of the
lower flight, and rtgt and sst each equal to a riser of the
upper flight. Draw gt s and gkt horizontal and equal
each to a tread of each flight respectively. Through r, draw
k, au, and through gy draw st tt. Vertically over d draw at tr
Horizontally draw atl anll and tt ttl.
It is usual to extend the wreath of the cylinder so as to
include a part of the straight rail — such a part as convenience
may require. Let the straight part here to be included ex-
tend from / to b on the plan. Vertically over b draw bt ctlli,
and horizontally draw b/ w-tl ; at any point on bt wtt locate wn,
and make wtl wf equal to j I, and bisect it in w; erect the
perpendiculars ivt altil, w dv/J, and w/7 v ; join tlt and atlll ; from
dvil horizontally draw dvil dv/ ; parallel with rt kt draw
dv, ctlll' We now have the plan and elevations of the prism,
FACE-MOULD FOR PLATFORM STAIRS.
265
containing at its angles the tangents required for the wreath
extending from b to d on the plan. The elevation F is a view
of the cylinder looking in the direction dc.
FIG. 142.
Comparing Fig. 142 with Fig. 141* the line b, w/t is the
trace, upon a vertical plane, of the horizontal plane abed
266 STAIRS.
of Fig. 141, or is the ground-line from which the heights of
the prism are to be taken.
The triangle^ bt au is represented in Fig. 141 at ab g, and
the inclined line b, atl is the tangent of the rail of the lower
flight, and is represented in Fig. 141 at bg ; while aniltn is
the tangent of the railing around the cylinder, and the half
of it is represented in Fig. 141 at ge. The height btctilJ is
shown in Fig. 141 at c f, while the height iv dv,t, or a/dVl, is
shown in Fig. 141 at de.
The vertical planes EEC may now be constructed about
the prism as in Fig. 140, proceeding thus :. Make c cf equal to
bt ciul, and ddt equal to at dv/ ; through ct draw di h\ through
b draw h btl ; perpendicular to h blt through a draw bn dv ;
from ^parallel with bn */vdraw d djiti ; on d as centre describe
the arc dtdnil\ draw ditll dv, also d dltl, parallel with hbn\
on dti as centre describe the arc dvdtj ; join dw to bjt. Par-
allel with bn h draw from each important point of the plan,
as shown, an ordinate extending to the line btl dtji, and thence
across plane D draw ordinates perpendicular to blt dtll, and
make them respectively equal to the corresponding ordinates
of the plane A, measured from the line bn dv; join e to/, all{
to b^ altl to e, and b, to/; also join /, to rt. Then ain b, is the
tangent standing over a b, and alit e is Jthe tangent standing
over ad. The line blli is the part of the tangent which
stands over blt, the portion of the wreath which is straight.
The curve enlplll is the trace upon the cutting plane of the
quarter circle dnpl, traced through the points /*,/,, and as
many more as desirable, found by ordinates as any other
point in the plane A. Thus we have complete the line
bt I, nt e, the central line of the wreath extending from b to d
in the plan. This is the essential part of the face-mould, which
is now to be drawn as follows: At Fig. 143 repeat the par-
allelogram atll btfe of Fig. 142, and, with a radius equal to
half the diameter of the rail, describe, from centres taken on
the central line, the several circles shown ; and tangent to
these circles draw the outer and inner edges of the rail.
The joint at bt is to be drawn perpendicular to the tangent
b. ain, while that at e is to be perpendicular to the tangent
^atll. This completes the face-mould for the wreath over
WREATHS FOR A ROUND RAIL. 267
bind of the plan. If the pitch-board of the upper flight be
the same as that of the lower flight, the face-mould at Fig.
143 will, reversed, serve also for the wreath over the other
half of the cylinder.
In using this face-mould, place it upon a plank equal in
thickness to the diameter of the rail, mark its form upon the
plank, and saw square through ; then chamfer the wreath to
an octagonal form, after which carefully remove the angles
so as to produce the required round form. The joints, as well
as the curved edges, are to be cut square through the plank.
Many more lines have been used in obtaining this face-
mould than were really necessary for so simple a case, but no
more than was deemed advisable in order properly to eluci-
date the general principles involved. A very simple method
FIG. 143.
for face-moulds of platform stairs with small cylinders will
now be shown.
269. — More Simple method for Hand-Rail to Platform
Stairs. — In Fig. 144, jge represents a pitch-board of the first
flight, and d and i the pitch-board of the second flight of a plat-
form stairs, the line e f being the top of the platform ; and
abc is the plan of a line passing through the centre of
the rail around the cylinder. Through i and d draw i k,
and through y and e drawy k ; from k draw k I parallel to/e;
from b draw bm parallel to gd ; from / draw Ir parallel to
kj ; from n draw nt at right Angles toy/6; on the line ob
make ot equal to nt ; join c and t ; on the line jc, Fig. 145,
make ec equal to en at Fig. 144; from c draw c t at right
angles to j c, and make ct equal to c t at Fig. 144; through /
draw p I parallel to j c, and make // equal to / / at Fig. 144 ;
join /and c, and complete the parallelogram eels ; find the
points o, o, o, according to Art. 551 ; upon e, o, o, o, and /,
268
STAIRS.
successively, with a radius equal to half the width of the
rail, describe the circles shown in the figure ; then a curve
traced on both sides of these circles, and just touching them,
FIG, 144.
Avill give the proper form for the mould,
drawn at right angles to c /.
The joint at / is
FIG. 145.
This simple method for obtaining the face-moulds for the
hand-rail of a platform stairs appeared first in the early edi-
tions of this work. It was invented by a Mr. Kells, an
HAND-RAIL TO PLATFORM STAIRS.
269
eminent stair-builder of this city. A comparison with Fig.
142 will explain the use of the few lines introduced. For a
full comprehension of it reference is made to Fig. 146, in
which the cylinder, for this purpose, is made rectangular
FIG. 146.
instead of circular. ,The figure gives a perspective view of
a part of the upper and of the lower flights, and a part of
the platform about the cylinder. The heavy lines, ////, me,
and cj, show the direction of the rail, and are supposed to
pass through the centre of it. Assuming that the rake of
270
STAIRS.
the second flight is the same as that of the first, as is gener-
ally the case, the- face-mould for the lower twist will, when
reversed, do for the upper flight ; that part of the rail, there-
fore, which passes from e to c, and from c to /, is all that will
need explanation.
Suppose, then, that the parallelogram eaoc represent a
plane lying perpendicularly over 'eabf, being inclined in
the direction ec, and level in the direction co ; suppose this
FIG. 147.
plane eaoc be revolved on ec as an axis, in the manner indi-
cated by the arcs o n and a x, until it coincides with the
plane ertc\ the line ao will then be represented by the line
x n ; then add the parallelogram xrtn, and the triangle ctl,
deducting the triangle ers\ then the edges of the plane cslc,
inclined in the direction ec, and also in the direction c I, will
lie perpendicularly over the plane eabf. From this we
gather that the line co, being at right angles to ec, must, -in
HAND-RAIL FOR LARGE CYLINDER. 2JI
order to reach the point /, be lengthened the distance nt,
and the right angle ect be made obtuse by the addition to
it of the angle tc /. By reference to Fig. 144, it will be seen
that this lengthening is performed by forming the right-
angled triangle cot, corresponding to the triangle cot in
Fig. 146. The line ct is then transferred to Fig. 145, and
placed at right angles tov^r; this angle ect is then increased
by adding the angle tcl, corresponding to tcl, Fig. 146.
Thus the point / is reached, and the proper position and
length of the lines ec and ^/obtained. To obtain the face-
mould for a rail over a cylindrical well-hole, the same process
is necessary to be followed until the length and position of
these lines are found ; then, by forming the parallelogram
eels, and describing a quarter of an ellipse therein, the
proper form will be given.
FIG. 148.
270. — Hand-Railling for a Larger Cylinder. — Fig. 147
represents a plan and a vertical section of a line passing
through the centre of the rail as before. From b draw bk
parallel to cd\ extend the lines zWandyV until they meet kb
in k and /; from ;/ draw nl parallel to ob; through / draw
// parallel to j k; from k draw kt at right angles to/£; on
the line ob make ot equal to kt. Make ec (Fig. 148) equal
to ek at Fig. 147 ; from c draw ct at right angles to ec, and
equal to ct at Fig. 147 ; from / draw // parallel to ce, and
make tl equal to //at Fig. 147 ; complete the parallelogram
eels, and find the points o, 0, o, as before ; then describe the
circles and complete the mould as in Fig. 145. The difference
between this and Case I is that the line ct, instead of being
raised and thrown out, is lowered and drawn in. A method
of planning a cylinder so as to avoid the necessity of cant-
ing the plank, either up or down, will now be shown.
2/2
STAIRS.
271. — Faee-moMld without Canting the Plank. — Instead
of placing the platform-risers at the spring of the cylinder, a
more easy and graceful appearance may be given to the
rail, and the necessity of canting either of the twists entirely
obviated, by fixing the place of the above risers at a certain
distance within the cylinder, as shown in Fig. 149 — the lines
indicating the face of the risers cutting the cylinder at k and
/, instead of at / and q, the spring of the cylinder. To
ascertain the position of the risers, let abc be the pitch-
board of the lower flight, and cde that of the upper flight,
these being placed so that b c and
cd shall form a right line. Extend
a c to cut de in f; draw fg parallel
*to db, and of indefinite length ;
draw go at right angles to fg, and
equal in length to the radius of the
circle formed by the centre of the
rail in passing around the cylinder ;
on o as centre describe the semi-
circle/^z'/ through o draw is par-
allel to db; make oh equal to the
radius of the cylinder, and describe
on o the face of the cylinder phq;
then extend db across the cylinder,
cutting it in / and k — giving the
position of the face of the risers,
as required. To find the face-
mould for the twists is simple and
obvious : it being merely a quarter
of an ellipse, having oj for semi-
minor axis, and sf for the semi-major axis; or, at Fig. 151,
let dci"be<\. right angle ; make c i equal to oj, Fig. 149, and dc
equal to sf, Fig. 149; then draw do parallel to ci, and com-
plete the curve as before.
272. — Railing for Platform Stair* where the Rake
meets the Level. — In Fig. 150, abc is the plan of a line pass-
ing through the centre of the rail around the cylinder as
before, and je is a vertical section of two steps starting
from the floor, kg. Bisect eh in d, and through d draw df
FIG. 149.
HAND-RAIL AT RAKE AND LEVEL.
273
parallel to hg\ bisect /# in /, and from / draw It parallel
to nj\ from n draw nt at right angles to jn ; on the line ob
make ot equal to nt. Then, to obtain a mould for the twist
going up the flight, proceed as at Fig. 145 ; making ec in
that figure equal to en in Fig. 150, and the other lines of a
length and position such as is indicated by the letters of
reference in each figure. To obtain the mould for the level
FIG. 150.
rail, extend bo (Fig. 150) to i ; make oi equal to //, and join
z'andc; vcw&e c i (Fig. 151) equal to civ&Fig. 150; thiough
FIG. 151.
c draw cd at right angles to ci\ make dc equal to df at
Fig. 150, and complete the parallelogram odd; then pro-
ceed as in the previous cases to find the mould.
273.— Application of Face-lWoiilds to Plank. — All the
moulds obtained by the preceding examples have been for
round rails. For these, the mould may be applied to a plank
of the same thickness as the rail is intended to be, and the
274
STAIRS.
plank sawed square through, the joints being cut square
from the face of the plank. A twist thus cut and truly
rounded will hang in a proper position over the plan, and
present a perfect and graceful wreath.
274. — Face-Moulds for Moulded Rails upon Platform
Stairs. — In Fig. i$2,abcis the plan of a line passing through
FIG. 152.
the centre of the rail around the cylinder, as before, and the
lines above it are a vertical section of steps, risers, and plat-
form, with the lines for the rail obtained as in Fig. 144. Set
half the width of the rail from b to / and from b to r, and
from / and r draw .fc and rd parallel to ca. At Fig. 153
the centre-lines of the rail jc and cl are obtained as in the
previous examples, making jc equal jn of Fig. 152, ct
FACE-MOULD APPLIED TO PLANK. 2/5
equal ct of Fig. 152, and tl equal si of Fig. 152. Make ci
and ck each equal to <:z at /^. 152, and draw the lines im
and /&£• parallel to cj ; make /<? and /r equal to ne and ?z</ at
7%. 152, and draw dn and eq parallel to lc\ also, through j
draw <?£• parallel to lc; then, in the parallelograms mnro
and goe q, find the elliptic curves, d??« and *£•, according to
Art. 551, and they will define the curves. The line dp,
being drawn through / perpendicular to Ic, defines the
joint which is to be cut square through the plank.
275. — Application of Face-Moulds to Plank. — In Fig.
152 make a drawing, from d to /*, of the cross-section of the
hand-rail, and tangent to the lower corner draw the line gh.
The distance between the lines/* and^/z is the thickness of
the plank from which the rail is to be cut. Lay the face1
mould upon the plank, mark its shape upon the plank, and
FIG. 153.
saw it square through. To proceed strictly in accordance
with the requirements of the principles upon which the face-
mould is formed, the cutting ought to be made vertically
through the plank, the latter being in the position which it
would occupy when upon the stairs. Formerly it was the
custom to cut it thus, with its long raking lines. But, owing
to the great labor and inconvenience of this method, efforts
were made to secure an easier process. By investigation it
was found that it was possible, without change in the face-
mould, to cut the plank square through and still obtain the
correct figure for the railing, and this method is the one now
usually pursued. Not only is the labor of sawing much re-
duced by this change ; but to the workman it is an entire re-
lief, as he now, after marking the form of the wreath upon the
plank, sends it to a steam saw-mill, and, at a small cost, has it
STAIRS.
cut out with an upright scroll-saw. When thus cut out in
the square, the upper surface of the plank is to be faced up
true and unwinding, and the outer edge jointed straight
and square from the face. Then a figure of the cross-section
of the hand-railing is to be carefully drawn on the ends of the
squared block as shown in Figs. 154 and 155, and which
are regulated so as to be correctly in position, as follows.
First, as to the end h of the straight part hj\ In Fig. 154,
let a b c d be an end view of the squared block, of which a e fd
is the shape of the end of the straight part. Let the point g
be the centre of this end of the straight part ; through g
draw upon the end a efd the line jk, so that the angle bjk
shall be equal to the angle kt c, Fig. 152. This is the angle
at which the plank is required to be canted, revolving it on
FIG. 154.
FIG. 155.
the axis of the straight part of the rail. Through g draw
the line ;/ h parallel with a b. Upon a thin sheet of metal
(zinc is preferable) mark carefully the exact figure of the
cross-section of the rail, drawing a vertical line through its
centre, cut away the surplus metal, then, with this template
as a pattern, mark upon the end a efd, Fig. \ 54, the figure of
the rail as shown, the vertical line upon the template being
made to coincide with the line//£. From n and h draw the
vertical lines // in and /;/ parallel with j k.
Now, as to the other end of the square block : Let b c f e,
Fig. 155, represent the block, of which bcvn is the form of
the end at the curved part, and o its centre. Through o
draiv/^, so that the angle epq shall be equal to the angle
j n /7, Fig. 152. Also, through o draw d h parallel with e b\
CUTTING THE TWIST-RAIL. 277
from d and h draw the vertical lines h r and ds parallel with
pq. Place the template on bcvn, the end of the block, so
that the vertical line through its centre shall coincide with
pq\ mark its form, then from y, at mid-thickness, draw wy
parallel with p q.
In applying- the mould, let Fig. 156 represent the upper
face of the squared block,
with the face-mould lying
upon it. With the distance
a /, Fig. 154, and by the
edge a x, mark a gauge-line
upon the upper face of the FIG. 156.
squared block. Set the outer edge of the face-mould to coin-
cide with this gauge-line. Let the end of the face-mould be
set at w, e w being equal to e w, Fig. 155; then mark the
block by the edge of the face-mould.
Now turn the block over and apply the face-mould to the
underside, as in Fig. 157. With the distance d r/tt Fig. 154,
and by the outer edge of
the block, mark a gauge-
line from m, Fig. 157. Set
the inner edge of the face-
mould to this gauge-line,
and slide it endwise till the FlG- J57.
distance em shall equal ew, Fig. 155, then mark the block by
the edges of the face-mould. The over wood may now be re-
moved as indicated by the vertical lines at the sides of the
cross-section marked on each end of the block (see also Fig.
167) : the direction of the cutting at the curves must be verti-
cal ; the inner curve will require a round-faced plane. A com-
parison of the several figures referred to, with the directions
given, together with a little reflection, will manifest the
reasons for the method here given for applying the face-
mould. Especially so when it is remembered that the face-
mould was obtained not for the top of the rail, but for the rail
at the mid-thickness of the block. So, therefore, in the
application to the upper surface of the block, the face-mould
is slid up the rake far enough to put the mould in position
vertically over its true positional mid-thickness ; and on the
2/8 STAIRS.
contrary, in applying the face-mould to the underside of the
plank, it is slid down until it is vertically beneath its true
position at the mid-thickness of the block.
When the vertical faces are completed, the over wood
above and below the wreath is to be removed. In doing
this, the form at the ends, as given by the template, is a suf-
ficient guide there. Between these the upper and under
surfaces are to be warped from one end to the other, so as
to form a graceful' curve. With a little practice an intelli-
gent mechanic will be able to work these surfaces with
facility. The form of cross-section produced by this opera-
tion is that of a parallelogram, tangent to the top, bottom,
and two sides of the rail ; and which at and near the ends
of the block is not quite full. The next operation is that of
working the moulding at the sides and on top, first by re-
bates at the sides, then chamfering, and finally moulding the
curves. Templates to fit the rail, one at the sides, another
on top, are useful as checks against cutting away too much
6f the wood. f
The joints are all to be worked square through the plank
in the line drawn perpendicular to the tangent, as shown in
153-
276. — Hand-Railing for Circular Stairs. — Let it be re-
quired to furnish the face-moulds for a circular stairs similar
to that shown in Fig. 133.
Preliminary to making the face-moulds it is requisite to
make a plan, or horizontal projection of the stairs, and on
this to locate the projections of the tangents and develop
their vertical projections. For this purpose let b c d e f g,
Fig. 158, be the horizontal projection of the centre of the
rail, and the lines numbered from i to 19 be the risers. At
any point, a, on an extension of the line of the first riser
locate the centre of the newel. On a as a centre describe the
two circles; the larger one equal in diameter to the diame-
ter of the newel-cap, the inner one distant from the outer
one equal to half the width of the rail. Let the first joint in
the hand-rail be located at b, at the fourth riser ; through b
draw h k tangent to the circle. Select a point, h, on this
PLAN OF CIRCULAR STAIRS.
279
tangent which shall be equally distant from b and from the
inner circle of the newel-cap, measured on a line tending to
a ; join h and a, and from a point, <?, on the line b o describe
ft,
10
FIG. 158.
the curve from b to the point of the mitre of the newel-cap,
the curve being tangent, at this point, to the line a h. Select
positions for the other joints in the hand-rail as at c, d, c, and /.
280 STAIRS.
Through these draw lines tangent to the circle.* Then the
horizontal projection of the tangents will be the lines hk, kl,
/;//, m n, and np. Now, if a vertical plane stand upon each
of these lines, these planes would form a prism not quite
complete standing upon the base-plane, A. Upon these ver-
tical planes, C, D, E, F, G, and H, lines may be drawn which
at each joint shall be tangent to the central line of the rail.
These are the tangents now to be sought. Perpendicular to
the tangents at £, c, d, etc., draw the lines b bt, c c^ ddjy e et, ff/y
ggn and h htl, kkt, k klt,l lt,l lin etc. As b is at the fourth riser,
and the height is counted from the top of the first riser,
make b bt equal to three risers. (To avoid extending the
drawing to inconvenient dimensions, the heights in it are
made only half their actual size. As this is done uniformly
throughout the drawing, this reduction will lead to no error
in the desired results.) As c is on the eighth riser, therefore
make c ct equal to seven risers, and so, in like manner, make
the heights ddltee^ and fft each of a height to correspond
with the number of the riser at which it is placed, deduct-
ing one riser. These heights fix the location of each tangent
at its point of contact with the central line of the rail. But
each tangent is yet free to revolve on this point of contact,
up or down, as may be required to bring the ends of each
pair of tangents in contact; or, to make equal in height the
edges of each pair of vertical planes, which coincide after
they are revolved on their base-lines into a vertical position ;
as, for example : the edges k kt and k kit of the planes C and
D must be of equal height; so, also, the edges llf and lltl of
the planes D and E must be of equal height. The method
of establishing these heights will now be shown.
To this end let it be observed, that of the horizontal pro-
jection of any pair of intersecting tangents, their lengths,
from the point of intersection to the points of contact with
the circle, are equal ; for example : of the two tangents // k
and Ik, the distances from k, their point of intersection, to b
and c, their points of contact with the circle, are equal ; and
so also cl equals <//, dm equals e m, etc. It will be observed
* A tangent is a line perpendicular to the radius, drawn from the point of
contact.
THE FALLING-MOULD FOR THE RAIL. 28 1
that this equality is not dependent on b,c, d, etc., the points
of contact, being disposed at equal distances ; for, in this
example, they are placed at unequal distances, some being at
three treads apart and others at four; and yet while this un-
equal distribution of the points b, c, d, etc., has the effect of
causing the point of contact, as b, c, or ey to divide each whole
tangent into two unequal parts, it does not disturb the
equality of the two adjoining parts of any two adjacent tan-
gents. Now, because of this equality of the two adjoining
parts of a pair of tangents, the height to be overcome in
passing from one point of contact to the next must be
divided equally between the two ; each tangent takes half
the distance. Therefore, for stairs of this kind, the arrange-
ment being symmetrical, we have this rule by which to fix
the height of the ends of any two adjoining tangents, namely :
To the height at the lower point of contact add half the dif-
ference between the heights at the two points of contact ;
the sum will be the required height of the two adjoining
ends of tangents. For example: the heights at b and c,
two adjacent points of contact, are respectively three and
seven risers; the difference is four risers; half this added to
three, the height of the lower rise, gives five risers as the
height of k kr kku, the height at the adjoining ends of the
tangents h k and / k. Again, the heights at c and d are re-
spectively seven and ten risers ; their difference is three ;
half of which, or one and a half 'risers, added to seven, the
height at the lower point of contact, makes nine and a half
risers as the heights //,, //„, at the ends of the adjoining
tangents k I and m I. In a similar manner are established
the heights of the tangents at ;;/, ;/, and /.
The rule for finding the heights of tangents as just given
is applicable to circular stairs in which the treads are di-
vided equally at the front-string, as in Fig. 158. Stairs of
irregular plan require to have drawn an elevation of the
rail, stretched out into a plane, upon which the tangents can
be located. This will be shown farther on.
The locations of the joints c, d, c, in this example, were
disposed at unequal distances merely to show the effect on
the tangents as before noticed. In practice it is proper to
282 STAIRS.
locate them at equal distances, for then one face-mould in
such a stairs will serve for each wreath.
When the tangent at G has been drawn, the level tangent
for the landing maybe obtained in this manner: As the
joint f is located at the eighteenth riser, one riser below the
landing, draw a horizontal line at s, one riser above the point
f^ and at half a riser above this draw the level line at pt \ then
this line is the level tangent, and p its point of intersection
with the raking tangent. Draw the vertical line /,/, and
from/ draw the tangent /£-, which is the horizontal projec-
tion of the tangent pt g, on plane H (which, to avoid undue
enlargement of the drawing, is reduced in height), where
////equals//,,.
To obtain the horizontal tangent / u at the newel, pro-
ceed thus : Fix the point r, in the tangent r kt, at a height
above b t equal to the elevation of the centre of the newel
above the height of a short baluster — for example, from 5
to 8 inches — and draw a line through r parallel to b t ; this
is a horizontal line through the middle of the height of the
newel-cap, and upon which and the rake the easement to
the newel is formed. Perpendicular to b t draw r t, and join
/ and u ; then / u is the horizontal tangent.
277.— Face-Moulds for Circular §tair§. — At Fig. 159 the
plan of the newel and the adjacent hand-rail are repeated,
but upon an enlarged scale ; and in which b bt is the reduced
height of the point &, or is equal to b bt less tr, Fig. 158,
and the angle b bt /equals the angle bbtr of Fig. 158. In
this plan the actual heights must now be taken. Join t and
u ; then / u is the level tangent, as also the line of intersection
of the cutting plane C and the horizontal plane A. Perpen-
dicular to / u, at a point / or anywhere above it, draw u, b/r
Parallel with / u draw b bltl ; make bn bltl equal to b bt ; join
bltl and u^ ; then the angle b bul ut is the angle which the
plank in position makes with a vertical line, or what is
usually termed the plumb-beviL Perpendicular to bltl uj
draw «, «„ and bltl bltll ; make btll btlll equal to bbtl ; make u,
tt equal to ut t, and uj un to u/ u ; join blltl and tt ; then bini tt is
the tangent in the cutting plane, the horizontal projection of
FACE-MOULD FOR FIRST SECTION.
283
which is bt. The butt-joint at bltil is drawn square to the
tangent bilu tr Parallel to the intersecting line / u, draw
ordinates across the plane A from as many points as desir-
able, and extend them to the rake-line ut bni ; through the
points of their intersection with this line, and perpendicular
to it, draw corresponding ordinates across the plane C. Make
du dtll equal to d, d, and so in like manner, for all other points,
FIG. 159.
obtain in the plane C for each point in the horizontal plane
A its corresponding point in the plane C: in each case
taking the distance to the point in the plane A from the line
h btl and applying it in the plane C from the rake-line u, bnl.
For the curves bend a flexible strip to coincide with the
several points obtained, and draw the curve by the side of
the strip. The point of the mitre is at */,„, the mitre-joint is
284 STAIRS.
shown at hdtll and dtllclt. The line f clt is drawn through
c//t the most projecting point of the mitre, and parallel to the
rake-line ut blir Additional wood is left attached, extending
from h to f\ this is an allowance to cover the mitre, which
has to be cut vertically ; the butt-joint at bltll and the face at
fctl are both to be cut square through the plank. The face
/<:„, because it is parallel to the rake-line uf bltl, is a vertical
face, as well as being perpendicular to the surface of the
plank. On it, therefore, lines drawn according to the rake,
or like the angle ut btll bit, will be vertical and will give the
direction of the mitre-faces. We now have at C the face-
mould for the railing over the plan from b to d in A. The
mould thus found is that made upon a cutting plane C, passed
through the plank, parallel to its face, but at the middle of
its thickness. To put it in position, let the plane C be lifted
by its upper edge ctl and revolved upon the line nt btll until
it stands perpendicular to the plane B. Now revolve both
C and B (kept in this relative position during the revolution)
upon the line ut bt. until the plane B stands perpendicular to
the plane A. Then every point upon plane C will be verti-
cally over its corresponding point in the plane A. For ex-
ample, the point blllt will be vertically over b, t, over /,
and so of all other points. To show the application of the
face-mould to the plank, make bltl bv equal to half the thick-
ness of the plank; parallel to ut btll draw bvc, a line which
represents the upper surface of the plank, for the line ut bul
is at the middle of the thickness. Through bitll, and parallel
with bltl ut, draw the line ct bltll and extend it across the face-
mould ; make blitl ct equal to bv c ; through c^ and parallel with
bull t/t draw ct e. Now, ;// n ot p is an end view of the plank,
showing the face view of the butt-joint at bi{ll. Through r,
the centre, draw a line parallel with the sides. Then #vl rep-
resents the point bltll\ make £vi et equal to btlil c\ through r,
the centre, draw c, r across the face of the joint ; then et r is
a vertical line (see Art. 284), parallel and perpendicular to
which the four sides of the squared-up wreath are to be
drawn as shown. In applying the face-mould to the plank at
first, for the purpose of marking by its edges the form of the
face-mould, it will be observed that the face-mould is under-
stood to have the position indicated by the line ut blu, or at
FACE-MOULDS FOR CIRCULAR STAIRS. 285
the middle of the thickness of the plank. By this marking
the rail-piece is cut square through the plank, and this cut-
ting gives the correct form of the wreath, but only at the
middle of the thickness of the plank. After it is cut square
through the plank, then, to obtain the form at the upper and
under surfaces, the face-mould is required to be moved end-
wise, but parallel with the auxiliary plane ^,-and so far as to
bring the face-mould into a position vertically over or under
its true position at the middle of the thickness of the plank.
For example, the point btlll^ if the mould were placed at the
middle of the thickness of the plank, would be at the height
of the point bnl; but when upon the top of the plank, the
point btlll would have to be at the height of the point ct,
therefore the mould must be so moved that the point blltl
shall pass from bv to c ; consequently bv c is the distance
the mould must be moved, or, as it is technically termed,
the sliding distance; hence blltl c,, which is equal to bv c, is
the distance the mould is to be moved: up when on top,
and down when underneath. This is more fully explained
in Art. 284.
278. — Face-Moulds for Circular Stairs. — At Fig. 1 60 so
much of the horizontal projection of the hand-railing of
stairs in Fig. 158 is repeated as extends from the joint b to
that at d, but at an enlarged scale. Upon the tangent ck
set up the heights as given in Fig. 158; for example, make
kkt equal to kniktl of Fig. 158, and cc, equal to cllcl of Fig.
158. Join ct and kt and extend the line to meet ck, extended,
in a. Join a and b ; then ab is the line of intersection of
the cutting and horizontal planes ; it is therefore a horizon-
tal line, parallel to which the ordinates are to be drawn.
Perpendicular to ab draw b,c,til. Parallel to ab draw ccu
and kklt ; join b, and clt ; the angle cctl bf is the plumb-bevil ;
perpendicular to b, cf. draw b, blt, ktl kin, and ctl cllt ; make bt bu
equal to bt b, and so of the other two points, kltl and r///} make
them respectively equal to their horizontal projections upon
the plane A. Join ciu and kfl ; also, kit and btl ; then btlkltl
and ktll cltl are the tangents. From t//f draw the line cltl blt
parallel to btctl ; this is the slide-line. In this example, this
286
STAIRS.
line passes through the point blt ; the slide-line does not
always pass through the ends of the two tangents ; it is not
required to pass through both, but it is indispensable that it
be drawn parallel with the rake-line bt ctl. The lines for the
joints at each end are drawn square to the tangent lines.
Points in the curves, as many as are desirable, are now to
be found by ordinates as shown in the figure, and as before
explained for the points in the tangents. The curves are
made by drawing a line against the side of a flexible strip
bent to coincide with the points.
The face-mould may be put in position by revolving the
planes C and B, as explained in the last article, for the rail
at the newel.
The face-mould for the rail over the plan from c to d is to
FACE-MOULDS CONTINUED. 287
be obtained in a similar manner, taking the heights from Fig.
158. For example, make d d/ equal to dtldt of Fig. 15.8, and
//, equal to /„,/„ of Fig. 158 (taking the heights at their
actual measurement now). Join dt and /„ and extend the
line to meet the line dl extended in r ; join r and c\ then re
is the line of intersection, and parallel to which the ordinates
are to be drawn. The points in the face-mould may now be
obtained as in the previous cases, giving attention first to
the tangent and slide-line ; drawing the lines for the joints
perpendicular to the tangents.
It may be remarked here that the chord-line £<:is parallel
with the measuring line btcilin and that the line o k bisects
the chord-line; so, also, the line ol bisects the chord-line cd.
This coincidence is not accidental ; it will always occur in a
regular circular stairs.
Hence in cases of this kind it is not necessary to go
through the preliminaries by which to obtain the intersect-
ing line ab, but draw it at once parallel to the line ok,
bisecting the chord be and passing through the point of
intersection of the two tangents! For the distance to slide
the mould in its after-application, the lines are given at ctl
and dlfj and their use is explained in the last article, and
more fully in Art. 284.
279. — Face-Mould§ for Circular Stairs again. — At Fig.
161 so much of the plan of the hand-railing of the stairs of
Fig. 158 is repeated as is required to show the rail from f
to gy but drawn at a larger scale. To prepare for the face-
moulds, perpendicular to // draw //., and make ppt equal
t° PuiPii °f F*8- !5^ (taking this height now at its actual
measurement) ; join pt and /; then fpt is the tangent of the
vertical plane C, and / is a point in the cutting plane at its
intersection with the bass-plane A. Now since rs, the tan-
gent over pg, is horizontal and is in the cutting plane,
therefore from / draw fa parallel with r s or pg\ then fa
is the line of intersection of the cutting and horizontal
planes, and gives direction to the ordinates. Draw //,„
perpendicular to fa ; make plllpjl equal to ppt ; join /„ and
/,; then the angle //„// is the plumb-bevil ; perpendicular
288
STAIRS.
to /„/, draw //„ and /„/„„ ; make /„/„„ equal to ptug,
pitd equal to plltp\ join d and /„ ; then *//„ and dplin are
the tangents. Make ptl e equal to half the thickness of the
plank ; draw ftl a parallel Avith ftplt ; make ftl a equal to e c ;
draw act parallel with the tangent fnd\ through /„, per-
pendicular to fti d, draw the line for the butt-joint ; then fu c,
is the distance required to determine the vertical line on the
face of the joint at /„, as shown at A. Through pltli, per-
pendicular to the tangent pull d, draw the line for the butt-
joint ; make pnil b equal to ec\ then plltlb is the distance
required for determining the vertical line on the face of the
joint at ///7/, as shown at B (see Art. 284). The curved lines
are obtained by drawing a line against, the edges of a flexi-
ble rod bent to as many points as desirable, obtained by
measuring the ordinates of the plan at A and transferring
them to the face-mould by the corresponding ordinates, as
before explained.
RAILING FOR QUARTER-CIRCLE STAIRS.
289
280.— Hand -Railing for Winding Stairs.— -The term
winding is applied more particularly to a stairs having steps
of parallel width compounded with those which taper in
width, as in Fig. 135, and as is here shown in Fig. 162, in
which fa be represents the central line of the rail around the
cylinder, and the quadrant de, distant from the first quadrant
20 inches, is the tread-line, upon which from d, a point taken
at pleasure, the treads are run off. Through e, perpendicu-
FIG. 162.
lar to af, draw ae (the occurrence here of one of the
points of division on the tread-line perpendicularly opposite
a, the spring of the circle, is only an accidental coincidence) ;
make a a, equal to two risers ; join a, and /. With the
diameter a c, on b as a centre, describe the arc at g, crossing
ac extended ; through b draw gb, ; then ab, is the stretch-
out, or development of the quadrant a b.
Through // draw h i, tending toward the centre of the
STAIRS.
cylinder; make btit equal to bi\ perpendicular to fbt draw
bt bit and it itl. As there are four risers from e to //, make
at ati equal to four risers, and draw atl iu parallel with fa ;
through iit draw at bi( ; by intersecting lines, or in any con-
venient manner, ease off to any extent the angle fat iu.
Through j, a point in this curve (chosen so as to be perpen-
dicularly over m, a point between a and /, nearer to a),
draw kly a tangent to the curve. Perpendicularly to this
tangent, through /, draw the line for a butt-joint ; also
through £//t and perpendicularly to a, blt, draw the line for
the joint at the centre of the half circle. On the line aan
set up points of division for the riser heights, and through
these points of division draw horizontal lines to the line
From these points of contact drop perpendiculars to the
line fa bt, and transfer such of them as require it to the circle
at, by drawing lines tending to g. Through these points of
intersection with the central line of the rail, and through the
points of division on the tread-line, draw the riser-lines me,
a n, etc. At half a riser above the floor-line, on top of the
upper riser draw a horizontal line, and ease off the angle as
shown ; the intersection of the floor-line with this curve
gives the position of the top riser at the centre of the rail.
This completes the plan of the steps and the elevation of the
rail — requisite preliminaries for the face-moulds. The gradu-
ation of the treads from flyers to winders obviates an abrupt
angle at their junction in the rail and front-string. The
objection to the graduation, that it interferes with the
regularity of stepping at the tread-line, is not realized in
practice.
281. — Face-Moulds for Winding Stairs. — At Fig. 163 so
much of the plan at Fig. 162 is repeated as is required for the
face-moulds, but for perspicuity at twice the size. The hori-
zontal projection of the tangents for the first wreath are ad
and db drawn at right angles to each other, tangent to the
circle at a and b. Let those tangents be extended beyond
d; through ;;/, the lower end of the wreath, draw mdjy mak-
ing an angle with md equal to that in Fig. 162, between the
FACE-MOULDS FOR THE TWISTS.
29I
line af and #,./; or let the angle dmdt equal a fa, of Fig.
162. Make ddlt equal to ddr Make bbtl equal to btllbtl of
Fig. 162 ; join dlt and btl and extend the line to e4l ; make
blt bv equal to bubllu of Fig. 162, and draw b^etl parallel with
FIG. 163.
</*. From ^/y draw etle parallel with bltb\ through e and /
draw ef tangent to the circle at /; then b e and ef are the
horizontal projections of the tangents for the upper wreath.
Then if the plane B be revolved on ad, the plane C on de,
2Q2 STAIRS.
and the plane D on cf until they each stand vertical to the
plane A, the lines mdn dtlein and enift will constitute the
tangents of the two wreaths in position. This arrangement
locates the upper joint of the upper wreath at /, leaving fc,
a part of the circle, to be worked as a part of the long level
rail on the landing. As the tangent over ef is level, the
raking part of the rail will all be included in the wreath bf,
so that at the joint / the rail terminates on the level.
The portion fc, therefore, is a level rail requiring no
canting, and it requires no other face-mould than that afforded
by the plan from / to c.
For the face-mould for the rail over ;;/ a b, let the line eff dtl
be extended to mv, a point in the base-line b mv ; then ;//v is a
point in the base-plane A, as well as in the cutting plane E;
therefore the line mv m is the intersecting line parallel to
which all the ordinates on plane A are to be drawn. Per-
pendicular to this intersecting line mvm,at any convenient
place draw m, b, ; make bt bni parallel to mv m and equal to
b btl ; connect bilt with ;//,, a point at the intersection of the
lines mv m and b, mt ; then the angle b bnl m/ is the plumb-
bevil. Through d, parallel to mv m, draw^^; from the
three points /#,, diti, and btil draw lines perpendicular to
mf bllt ; make mt mtl equal to mt m ; make bitl bllu equal to bt b.
Since the measuring base-line m, b, passes through d, the
point of the angle formed by the two tangents, dtil is the
point of this angle in the cutting plane E\ therefore join mtl
and </,„, also dltl and blltl ; then bltll dni and dltl mn are .the
two tangents at right angles to which the joints at mn and
blin are drawn. The curves of the face-mould are now found
as usual, by transferring the distances by ordinates, as shown,
from the plane A to the plane E, making the distance from
the rake-line m/ bltl to each point in plane E equal to the dis-
tance from the corresponding point in the plane A to the
measuring base-line m, bt. Now, to obtain the sliding distance
and the vertical line upon the butt-joints, make btll bv equal
to half the thickness of the plank ; parallel with mt btll draw
by b^ ; also, bllu £vii and m4l mtll ; make btlll bvYi and «*„ mllt
each equal to bv £vi ; through ^vii and mn/, and parallel to the
respective tangents, draw £vil b^ and mllt mliu ; then b* and
FACE-MOULDS FOR WINDING STAIRS. 293
m,ni are the points from which, through the centre of the
butt-joints, a line is to be drawn which will be vertical when
the wreath is in position. (See Art. 284.)
For the face-mould for the upper quarter, through b, Fig.
163, draw b et parallel with du etl ; make e elit equal to e et ;
draw elilfl parallel with e f. Now, since etll ft is a horizon-
tal line and is in the cutting plane F, therefore, parallel with
etilft and through btJ draw b n ; then b n is the required in-
tersecting line. Extend e f to /„ ; make //„ equal to //, ;
join/,, and n ; then the angle ffl{ n is the plumb-bevil. Per-
pendicular to nft/ draw /"„/"„, and n n,, and make these lines
respectively equal to e f and b n ; join ftl and fltl ; also /„,
and nt ; then fn ftil and fnl n{ are the required tangents.
The butt-joints at/) and nt are drawn perpendicular to their
respective tangents. To get the slide distance and vertical
lines on the butt-joints, make/), fv equal to half the thickness
of the plank ; parallel with n /),, through /v draw/v/)yy/ ; also,
through nt draw nt nti ; make n, ii'u equal to /v fntl ; through
»/;, parallel with ntfltlJ draw nu nlt.\ then nln is the point
through which a line is to be drawn to the centre of the
butt-joint, and this line will be in the vertical plane contain-
ing the tangent. So, also, parallel with the tangent ftl fltl,
and through ////y, draw flllt /vi ; then/vi is the point through
which a line is to be drawn to the centre of the butt-joint
(see Art. 284). The curve is now to be obtained by the
ordinates, as before explained.
282. — Face-]Woiilci§ for Winding Stairs, again. — In the
last article, in getting the face-moulds for a winding stairs,
the two wreaths are found to be very dissimilar in. length.
This dissimilarity may be obviated by a judicious location of
the butt-joint connecting the two wreaths, as shown in Fig.
164. Instead of locating the joint precisely at the middle of
the half circle, as was done in Fig. 163, place it farther down,
say at ;/, which is at n in Fig. 162, two risers down from the
top, or at any other point at will. Then through n in the
plan draw mt s tangent to the circle at n ; and perpendicu-
lar to this tangent draw ;/ nin and ddn ; make n ntl equal to
nt n of Fig. 162; from d erect d dt perpendicular to m d\
294
STAIRS.
make the angle d m dt equal to that of bin j I of Fig. 162.
Make d dn equal to d dt ; join dn and nn and extend the line
to ;;/,, a point of intersection with the base-line n nt ; then nt
is a point in the base-plane, as also in the cutting plane ;
FIG. 164.
therefore mt m is the intersecting line parallel to which all
the ordinates of the plan are to be drawn, and perpendicular
to which m/{ nt, the measuring base-line, is drawn. Make
nt ntlil equal to n nn ; connect mtl and n/(/i, and then transfer
CARE REQUIRED IN DRAWING. 295
by the ordinates to the cutting plane m d and n the three
points of the plan at the ends of the tangents, as before de-
scribed, as also such points in the curve as may be required
to mark the curve upon the face-mould, all as shown in previ-
ous examples. For the face-mould of the upper wreath, make
ntl nnl equal to n nn of Fig. 162. From ntll draw nnl stl par-
allel with mt s ; extend the line dlt nlt to intersect ntll su in slt ;
parallel with niu n draw s/f s ; from s draw s r tangent to the
circle at r (s n equals s r) ; through r, tending to the centre of
the cylinder, draw the butt-joint ; then r s and s n are the
horizontal projections of the tangents for the upper wreath-
piece, the tangent s r being level and, consequently, parallel
to the intersecting line drawn through n. Perpendicular to
r s draw rtp ; parallel with nu sit draw nst ; make rt rlt equal
toss,; join rn and /. From this line and the measuring base-
line r, py the points for the tangents are first to be obtained
and then the points in the curve, all as before described.
The part of the circle from r to c is on the level, as before
shown, and may be worked upon the end of the long level
rail, its form being just what is shown in the plan from c to r.
283. — Face-Moiild§ : Te§t of Accuracy. — The methods
which have been advanced for obtaining face-moulds are
based upon principles of such undoubted correctness that
there can be no question as to the results, when the methods
given are thoroughly followed. And yet, notwithstanding
the correctness of the system and its thorough comprehen-
sion by the stair-builder, he will fail of success unless he
exercises the greatest care in getting his dimensions, his per-
pendiculars, and his angles. The slightest deviation in a
perpendicular terminated by an oblique line will result in a
magnified error at the oblique line. To secure the greatest
possible degree of accuracy, care must be exercised in the
choice of the instruments by which the drawings are to be
made : care to know that a straight-edge is what it purports
to be ; that a square, or right-angle, is truly a right-angle ;
that the compasses or dividers be well made, the joint per-
fect, and the ends neatly ground to a point. Then let the
drawing-board be carefully planed to a true surface ; and,
296 STAIRS.
if possible, let the drawing, full size, be made upon large,
stout roll-paper rather than upon the drawing-board itself,
as then the points for the face-mould may be pricked through
upon the board out of which the face-mould is to be cut, and
thus a correct transfer be made. For long straight lines it
is better to use a fine chalk-line than the edge of a wooden
straight-edge. The line is more trustworthy. Perpendicu-
lars, especially when long, are better obtained by measure-
ment or by calculation (Art. 503) than by a square. The
pencil used should be of fine quality — rather hard, in order
that its point may be kept fine. With these precautions in
regard to the instruments used, and with due care in the
manipulations, the face-moulds may be correctly drawn,
accurate in size and form. As a test of the accuracy of the
work, it will be well to observe in regard to the tangents,
that the length of a tangent, as found upon the face-mould,
should always equal its length as shown upon the vertical
plane. For example, in Fig. 160, the tangent klt ctil on the
face-mould should be equal to kt ct, the tangent on the vertical
plane B ; and in cases like this, where the stairs are quite
regular, with equal treads at the front-string, the two tan-
gents of a face-mould are-equal to each other, or klf cin equals
ktl btl ; and in this case, the line btl cllt should equal the rake-
line b, *„.
Again, as another example, in Fig. 161, d fin the tangent
upon the face-mould, should be equal to//,, the tangent of
the vertical plane C\ while d ///;/, the other tangent on the
face-mould, should be equal to r s, the tangent of the vertical
plane D. But the more important test is in the length of the
chord-line joining the ends of the two tangents ; as, for ex-
ample, the chord mtl binl of Fig. 163, the horizontal projec-
tion of which is the chord ;;/ b in plane A. Perpendicular to
m b draw b g\ make b g equal to b blt, and join g and m ; then
m^ blilt, the chord of the face-mould, should be equal to ;// g.
After fully testing the accuracy of the drawing for the face-
mould, choose a well-seasoned thin piece of white-wood, or
any other wood not liable to split, and plane it to an even
thickness throughout ; mark upon it the curves, joints, tan-
gents, and slide-line, and cut the edges true to the curve-
THE FACE-MOULD APPLIED TO PLANK. 297
lines and joints square through the board ; then square over
such marks as are required to draw each tangent and the
slide-line also upon the reverse side of the board. This
completes the face-mould.
284. — Application of the Faee-UIoulcl. — In order that a
more comprehensive idea of the lines given for applying a
face-mould may be had, let A, Fig. 165, represent one end of
a wreath-piece as it appears when first cut from a plank, and
when held up in the position it is to occupy at completion
over the stairs. Also, let B represent the corresponding
face-mould, laid upon the wreath-piece A in the position
which it should have after sliding. And, for the purpose of
a clearer illustration, let it be supposed that the two pieces,
A and B, are transparent. Then let at a b d c^ et represent a
solid of wedge form, having a triangular level base, a b d,
upon the three lines of which stand these three vertical
planes, namely : on the line a b the plane at a b c^ upon the
line a d the plane a, a d et, and on the line d b the plane b d
et c, ; the top of the solid is an inclined plane, ai ct en and the
vertical line at a is the edge of the wedge. Now, it will be
observed that the point a in the base of the solid is identical
with a, the centre of the butt-joint, and the point a, (at the
intersection of two vertical planes and the inclined plane of
the solid) is vertically over a, and is identical with a,, a poim
in the upper surface of the plank. Also, the inclined plane
e/ c/ at, which forms the top of the solid, coincides with the
upper surface of the plank A, from which the wreath-piece
has been squared ; and the line c, at (at the angle formed by
the inclined plane e, c, a, and the vertical plane af a b c,) coin-
cides with / g, the slide-line drawn upon the top of the
plank ; also, the line e, a, (at the angle formed by the in-
clined plane e, c, at and the vertical plane at a d e^ coincides
with at k, the tangent line upon the underside of the face-
mould after it has been slid to its new position, vertically
over its true position at the middle of the thickness of the
plank. From a the line a c is drawn parallel with a, ct ; so,
also, the line a e is drawn parallel with a, e, ; consequently
the line e c is parallel with e, ct ; and the plane e c a is parallel
298
STAIRS.
with the plane e{ cf at, and coincides with a plane passing
through the middle of the thickness of the plank, and, conse-
quently, is the cutting plane referred to in previous articles,
upon which the lines are drawn which give shape to the
FIG. 165.
face-mould. When the face-mould is first laid upon the plank,
the line i,j, coincides with ilt j/t, and Hvhen in that position,
its form marked upon the plank is the form by which the
plank is sawed square through ; but this gives the form of
THE SLIDING OF THE FACE-MOULD. 299
the wreath, not as it is at the surface of the plank, but as it
is at the middle of the thickness of the plank, or upon the
plane ace-, so that, for example, the line iitjtl represents the
line ij drawn through a, the centre of the butt-joint ; and
when the mould B is slid to the position shown in the figure,
the line iljl comes into a position vertically over ij\ hence
the three lines if i, at a, and jtj are each vertical and in a
vertical plane, Hjj^j- By these considerations it will be seen
that the face-mould B, located as shown in the figure, is in its
true position for the second marking, by which the addi-
tional cutting is now to be performed vertically. This being
established, it will now be shown how to get upon the butt-
joint a line in the vertical plane containing the tangent. If
the top and bottom lines of the vertical plane at a b ct be ex-
tended, they will meet in the point /, and will extend the
plane into a triangle Ib ct, cutting the upper edge of the
butt-joint in/, the end of the tangent, and the point in which
the point at of the underside of the face-mould was located
when the mould was first applied to the plank. The line/Vz
on the butt-joint is perpendicular to i j or ilt ju. Again, if
the top and bottom lines of the plane at a d e, be extended,
they will meet in /, and will extend the plane into the tri-
angle p d et, cutting the edge of the butt-joint in h, a point
from which, if a line be drawn upon the butt-joint to a, its
centre, this line will be in the vertical plane pdet, which
plane contains the tangent perpendicular to which the butt-
joint is drawn ; consequently lines upon the butt-joint par-
allel to h a will each be in a vertical plane parallel to the
vertical tangent plane, and lines drawn upon the butt-joint
perpendicular to these lines will be horizontal lines ; hence
the line h a is the required line by which to square the
wreath at the butt-joint. Now, it will be observed that the
triangle af a, is like that given in the various figures for obr
taining face-moulds, to regulate the sliding of the face-mould
and the squaring at the butt-joint. For example, in Fig.
163, the right-angled triangle bul £v £vi is the one referred to.
This triangle is in a vertical plane parallel to one containing
the slide-line; its longer side is a vertical line ; one of the
sides containing the right angle is equal to half the thickness
300
STAIRS.
of the plank, while the other, drawn parallel to the face of
the plank, is the distance the face-mould is required to slide.
Precisely like this, the triangle a f at of Fig. 165 is in the
vertical plane / b cn containing fg, the slide-line ; its longer
side, a, a, is a vertical line ; fa, one of the sides containing
the right angle, is equal to half the thickness of the plank,
while the other side, drawn coincident with the surface of
FIG. 166,
the plank, is the distance to slide the face-mould. Therefore
the triangle atf a of Fig. 165 gives the required lines by
which to regulate the application of the face-moulds. The
relative position of the more important of these lines is geo-
metrically shown in Fig. 166, in which A and B are upon the
horizontal plane of the paper, C is in a vertical plane stand-
ing on the ground-line b d, and D is a plan of the butt-joint,
revolved upon the line itt jn into the horizontal plane, and
BLOCKING-OUT OF THE RAIL. 3<DI
then perpendicularly removed to the distance //,. The let-
tering corresponds with that in Fig. 165. The shaded part
of D shows the end of the squared wreath. When the
blocked piece has been marked by the face-mould in its
second application, its edges are to be trimmed vertically as
shown in Fig. 167, after which the top and bottom surfaces
of the wreath are to be formed from the shape marked on
the butt-joints.
FIG. 167.
285.— Face-Mould Curves are Elliptical.— The curves of
the face-mould for the hand-railing of any stairs of circular
plan are elliptical, and may be drawn by a trammel, or
in anv other convenient manner. The trouble, however,
attending the process of obtaining the axes, so as to be able
to employ the trammel in describing the curves, is, in many
cases, greater than it would be to obtain the curves through
points found by ordinates, in the usual manner. But as
302
STAIRS.
this method for certain reasons may be preferred by some,
an example is here given in which the curves are drawn by
a trammel, and in which the method of obtaining the axes is
shown.
Let Fig. 168 represent the plan of a hand-rail around part
FIG. 168.
of a cylinder and with the heights set up, the intersection
line obtained, the measuring base-line drawn, the rake-line
established, and the tangents on the face-mould located — all
in the usual manner as hereinbefore described. Then, to
prepare for the trammel, from o, the centre of the cylinder,
draw o bt parallel with the intersecting line, and bt ot perpen-
FACE-MOULDS FOR ROUND RAILS. 303
dicular to •£,/,, the rake-line ; make btot equal to bo, and olal
equal to oa; through ot draw oth parallel with btfr From
o draw oe perpendicular toob,; continue the central circular
line of the rail around to e; parallel with obt draw e fy
and from /„ the point of intersection of ef with btft, and
perpendicular to #,/,, draw ftet\ make flel equal to fe\
then 0, is the centre of the ellipse, and ot a, the semi-conju-
gate diameter and ot e, the semi-transverse diameter of an
ellipse drawn through the centre of the face-mould. To get
the diameters for the edges of the face-mould, make a/ c, and
a/di each equal to half the width of the rail, as at cad\ par-
allel to a line drawn from at to et, and through cr draw the
line ctg\ also, parallel with a line drawn from at to et draw
dth (see Art. 559); then for the curve at the inner edge of
the face-mould, otg is the semi-transverse diameter, and olcl
the semi-conjugate ; while for the curve at the outer edge
o.h is the semi-transverse diameter, and otdt the semi-conju-
gate. So much of the edges of the face-mould as are straight
are parallel with the tangent. Now, placing the trammel at
the centre, as shown in the figure, and making the distance
on the rod from the pencil to the first pin equal to the
semi-conjugate diameter, and the distance to the second pin
equal to the semi-transverse diameter, each curve may be
drawn as shown. (See Art. 549.)
286.— Face-Moulds for Round Hails. — The previous ex-
amples given for finding face-moulds are intended for moulded
rails. For round rails the same process is to be followed,
with this difference : that instead of finding curves on the
face-mould for the sides of the rail, find one for a centre-line
and describe circles upon it, as at Fig. 145 ; then trace the
sides of the mould by the' points so found. The thickness of
stuff for the twists of a round rail is the same as for the
straight part. The twists are to be sawed square through.
287. — Position of the Butt -Joint. — When a block for
the wreath of a hand-rail is sawed square through the
plank, the joint, in all cases, is to be laid on the face-mould
square to the tangent and cut square through the plank.
304
STAIRS.
Managed in this way, the butt-joint is in a plane pierced
perpendicularly by the tangent. But if the block be not
sawed square through, but vertically from the edges of the
FIG. 169.
face-mould, then, especially, care is required in locating the
joint. The method of sawing square through is attended
with so many advantages that it is now generally followed ;
yet, as it is possible that for certain reasons some may prefer,
rv
[(UNI VET? si
POSITION OF THE BUTT-JOINT." 305
in some cases, to saw vertically, it is proper that the method
of finding the position of the joint for that purpose should
be given. Therefore, let A, Fig. 169, be the plan of the rail,
and B the elevation, showing its side; in which kz is the
direction of the butt-joint. From k draw kb parallel to /<?,
and ke at right angles to kb\ from b draw b f, tending to
the centre of the plan, and from / draw fe parallel to bk;
from /, through e, draw I i, and from i draw id parallel to
ef-, join dfand b, and db will be the proper direction for the
joint on the plan. The direction of the joint on the other
side, a c, can be found by transferring the distances x b and
od to xa and oc. Then the allowance for over wood to
cover the butt-joint is shown as that which is included be-
tween the lines ox and db. The face-mould must be so
drawn as to cover the plan to the line b d for the wreath at
the left, and to the line a c for that at the right. By some
the direction of the joint is made to radiate toward the
centre of the cylinder ; indeed, even Mr. Nicholson, in his
Carpenter s Guide, so advised. That this is an error may be
shown as follows : In Fig. 170, arji is the plan of a part of the
rail about the joint, s u is the stretch-out of a i, and gp is the
helinet, or vertical projection of the plan arji. This is
found by drawing a horizontal line from the height set upon
each perpendicular standing upon the stretch-out line su.
The lines upon the plan arji are drawn radiating to the
centre of the cylinder, and therefore correspond to the
horizontal lines of the helinet drawn upon its upper and
under surfaces.
Bisect rt on the ordinate drawn from the centre of the
plan, and through the middle draw cb at right angles to gv ;
from b and c draw cd and be at right angles to su ; from d
and e draw lines radiating toward the centre of the plan ;
then do and em will be the direction of the joint on the
plan, according to Nicholson, and cb its direction on the
falling-mould. It must be admitted that all the lines on the
upper or the lower side of the rail which radiate toward
the centre of the cylinder, as do, cm, or ij\ are level; for
instance, the level line wv, on the top of the rail in the
helinet, is a true representation of the radiating line// on
306
STAIRS.
the plan. The line bh, therefore, on the top of the rail in
the helinet, is a true representation of e m on the plan, and
kc on the bottom of the rail truly represents do. From k
draw £/ parallel to cb, and from h draw hf parallel to bc\
FIG. 170.
join / and b, also c and /; then cklb will be a true repre-
sentation of the end of the lower piece, B, and cfh b of the
end of the upper piece, A ; and fk or hi will show how
much the joint is open on the inner, or concave, side of the
rail.
CORRECT LINES FOR BUTT-JOINT.
307
To show that the process followed in Art. 287 is correct,
let do and em (Fig. 171) be the direction of the butt-joint
found as at Fig. 169. Now, to project, on the top of the rail
in the helinet, a line that does not radiate toward the centre
of the cylinder, as jk, draw vertical lines from j and k to w
and A, and join w and h ; then it will be evident that wh is
a true representation in the helinet of jk on the plan, it
being in the same plane as j k, and also in the same winding-
surface as wv. The line /», also, is a true representation on
308
STAIRS.
the bottom of the helinet of the line jk in the plan. The
line of the joint e m, therefore, is projected in the same way,
and truly, by ib on the top of the helmet, and the line do
by ca on the bottom. Join a and i, and then it will be seen
that the lines c a, a t, and ib exactly coincide with cb, the line
of the joint on the convex side of the rail ; thus proving the
lower end of the upper piece, A, and the upper end of the
lower piece, B, to be in one and the same plane, and that the
direction of the joint on the plan is the true one. By refer-
ence to Fig. 169 it will be seen that the line li corresponds
to xi in Fig. 171 ; and that e k in that figure is a representa-
tion of fb, and ik of db.
288. — Scrolls for Hand-Rail§ : General Rule for Size
and Position of the Regulating Square. — The breadth
which the scroll is to occupy, the number of its revolutions,
and the relative size of the regulating square to the eye of
the scroll being given, multiply the number of revolutions
by 4, and to the product add the number of times a side of
the square is contained in the diameter of the eye, and the
sum will be the number of equal parts into which the breadth
is to be divided. Make a side of the regulating square
equal to one of these parts. To the breadth of the scroll
add one of the parts thus found, and half the sum will be the
length of the longest ordinate.
FIG. 172.
289. — Centres in Regulating Square. — Let a 2 I b (Fig.
T72) be the size of a regulating square, found according to
the previous rule, the required number of revolutions being
SCROLLS AT NEWEL.
309
if. Divide two adjacent sides, as a 2 and 2 i, into as many
equal parts as there are quarters in the number of revolu-
tions, as seven ; from those points of division draw lines
across the square at right angles to the lines divided ; then
i being the first centre, 2, 3,4, 5, 6, and 7 are the centres for
the other quarters, and 8 is the centre for the eye ; the heavy
lines that determine these centres being each one part less
in length than its preceding line.
FIG. 173-
290.— Scroll for Hand-Rail Over Curtail Step. — Let
a b (Fig. 173) be the given breadth, if the given number of
revolutions, and let the relative size of the regulating square
to the eye be \ of the diameter of the eye. Then, by the
rule, if multipled by 4 gives 7, and 3, the number of times a
side of the square is contained in the eye, being added, the
sum is 10. Divide a b, therefore, into 10 equal parts, and set
one from b to c ; bisect a c in e ; then a e will be the length
of the longest ordinate (i d or i e). From a draw a d,
from e draw e i, and from b draw b f, all at right angles to
a b ; make e \ equal to a, and through i draw i d parallel
310 STAIRS.
to a b ; set b c from i to 2, and upon i 2 complete the regu
lating square; divide this square as at Fig. 172; then de-
scribe the arcs that compose the scroll, as follows : upon i
describe d e, upon 2 describe e /, upon 3 describe / g,
upon 4 describe £-//, etc.; make dl equal to the width of the
rail, and upon i describe / ;//, upon 2 describe mn, etc.; de-
scribe the eye upon 8, and the scroll is completed.
291. — Scroll for Curtail Step. — Bisect d I (Fig. 173) in <?,
and make o v equal to £ of the diameter of a baluster ; make
v w equal to the projection of the nosing, and e x equal to
wl\ upon i describe ivy, and upon 2 describe y z ; also, upon
2 describe x i, upon 3 describe ij\ and so around to z ; and
the scroll for the step will be completed.
292. — Position of Baluster* Under Scroll. — Bisect dl
(Fig. 173) in 0, and upon i, with i o for radius, describe the
circle o r u ; set the baluster at p fair with the face of the
second riser, r2, and from/, with half the tread in the divi-
ders, space off as at o, q, r, s, /, u, etc., as far as f ; upon 2, 3,4,
and 5 describe the centre-line of the rail around to the eye
of the scroll ; from the points of division in the circle o r u
draw lines to the centre-line of the rail, tending to 8, the
centre of the eye ; then the intersection of these radiating
lines with the centre-line of the rail will determine the posi-
tion of the balusters, as shown in the figure.
293. — Fal ling-Mould for Raking Part of Scroll. — Tangi-
cal to the rail at h (Fig. 173) draw h k parallel to da ; then
ha* will be the joint between the twist and the other part
of the scroll. Make de* equal to the stretch-out of d e, and
upon d e* find the position of the point k, as at k* ; at Fig.
174, make e d equal to ^2 d in Fig. 173, and d c equal to d c*
in that figure ; from c draw c a at right angles to e c, and
equal to one rise ; make c b equal to one tread, and from b,
through a, draw bj\ bisect a c in /, and through / draw m q
parallel to eh', m q is the height of the level part of a
scroll, which should always be about 3^ feet from the floor ;
ease off the angle mfj, according to Art. 521, and draw
FACE-MOULD FOR THE SCROLL.
gw n parallel to' m x j, and. at a distance equal to the thick-
ness of the rail ; at a convenient place for the joint, as *,
draw in at right angles to bj\ through n draw/// at right
angles to e h ; make dk equal to d k~ in Fig. 173, and from k
draw ko at right angles toe A; at Fig. 173, make <//;2 equal
to d h in ^. 174, and draw //2 £2 at right angles to
FIG. 174.
then k a* and /i2 b* will be the position of the joints on the
plan, and, at Fig. 174, op and in their position on the falling-
mould ; and po in (Fig. 174) will be the required falling-
mould which is to be bent upon the vertical surface from h 2
to k (Fig. 173).
FIG. 175.
294.— Face-Mould for the Scroll.— At Fig. 173, from k
draw kr* at right angles to r2 d- at Fig. 172, make h r
equal to h* r* in Fig. 173, and from r draw r s at right
angles to rh ; from the intersection of r s with the level line
m q, through /, draw s t ; at Fig. 173, make //2 b* equal to ^ /•
in Fig. 172, and join b* and r2; from rt2, and from as many
3I2
STAIRS.
other points in the arcs, a * I and k d, as is thought neces-
sary, draw ordinates to r2 d at right angles to the latter;
make r b (Fig. 175) equal in its length and in its divisions to
the line r2 d2 in Fig. 173 ; from r, n, o, p, q, and / draw the
lines r k, n d, o a, p e, q f, and / c at right angles to r b, and
equal to r2 k, d* s-,f* a", etc., in Fig. 173 ; through the points
thus found trace the curves k I and a c, and complete the
face-mould, as shown in the figure. This mould is to be ap-
plied to a square-edged plank, with the edge, / b parallel to
the edge of the plank. The rake-lines upon the edge of the
plank are to be made to correspond to the angle s t h in
Fig. 174. The thickness of stuff required for this mould is
shown at Fig. 174, between the lines s t and u v — u v being
drawn parallel to s t.
f 7'
295. — Form of Newel-Cap from a Section of the Rail.
— Draw a b (Fig. 176) through the widest part of the given
section, and parallel to c d\ bisect a b in e, and through a, e,
and b draw h i, f g, and k j at right angles to a b ; at a con-
BORING FOR BALUSTERS.
313
venient place on the line fg, as o, with a radius equal to
half the width of the cap, describe the circle ijg\ make r I
equal to e b or e a ; join / and /, also / and i\ from the curve
/ b to the line / j draw as many ordinates as is thought
necessary parallel to fg' from the points at which these
ordinates meet the line I j, and upon the centre, 0, describe
arcs in continuation to meet op ; from n t x, etc., draw ns,tu,
etc., parallel tofg; make n s, t u, etc., equal to ef, wv, etc.;
make x y, etc., equal to z d, etc.; make o 2, o 3, etc., equal to
o n, o t, etc.; make 2 4 equal to n s, and in this way find the
length of the lines crossing o m ; through the points thus
found describe the section of the newel-cap as shown in
the figure.
FIG. 177.
296.— Boring for Balusters in a Round Rail before it
is Rounded. — Make the angle oct (Fig. 177) equal to the
angle o c t at Fig. 144; upon c describe a circle with a
radius equal to half the thickness of the rail ; draw the tan-
gent b d parallel to / c, and complete the rectangle e b d f,
having sides tangical to the circle ; from c draw c a at right
angles to o c; then, b d being the bottom of the rail, set a
gauge from b to a, and run it the whole length of the stuff ;
in boring, place the centre of the bit in the gauge-mark at a,
and bore in the direction a c. To do this easily, make chucks
as represented in the figure, the bottom edge,^/*, being par-
allel to o c, and having a place sawed out, as e f, to receive
the rail. These being nailed to the bench, the rail will be
held steadily in its proper place for boring vertically. The
distance apart that the balusters require to be, on the under-
side of the rail, is one half the length of the rake-side of the
pitch-board.
STAIRS.
SPLAYED WORK.
297. — The Bevels in Splayed Work. — The principles
employed in finding the lines in stairs are nearly allied to
those required to find the bevels for splayed work — such as
hoppers, bread-trays, etc. A method by which these may be
FIG. 178.
obtained will, therefore, here be shown. In Fig. 178, a b c is
the angle at which the work is splayed, and b d, on the
upper edge of the board, is at right angles to a b ; make the
angle f g.j equal to a b c, and from f draw f h parallel to
e a ; from b draw b o at right angles to a b ; through o draw
ie parallel to c b, and join e and d\ then the angle a e d will
be the proper bevil for the ends from the inside, or k d e
from the outside. If a mitre-joint is required, set f g, the
thickness of the stuff on the level, from e to ;«, and join m
and d-, then kdm will be the proper bevil for a mitre-joint.
If the upper edge of the splayed work is to be bevelled,
so as to be horizontal when the work is placed in its proper
position, then f gj, the same as a b c, will be the proper
bevel for that purpose. Suppose, therefore, that a piece in-
dicated by the lines kg, g f, and f h were taken off ; then a
line drawn upon the bevelled surface from d at right angles
to k d would show the true position of the joint, because it
would be in the direction of the board for the other side ;
but a line so drawn would pass through the point o, thus
proving the principle correct. So, if a line were drawn upon
the bevelled surface from d at an angle of 45 degrees to k d,
it would pass through the point n.
VIEW IN THE ALHAMBRA.
SECTION IV.— DOORS AND WINDOWS.
DOORS.
298. — General Requirements. — Among the architectural
arrangements of an edifice, the door is by no means the least
in importance ; and if properly constructed, it is not only
an article of use, but also of ornament, adding materially to
the regularity and elegance of the apartments. The dimen-
sions and style of finish of a door should be in accordance
with the size and style of the building, or the apartment
for which it is designed. As regards the utility of doors,
the principal door to a public building should be of suffi-
cient width to admit of a free passage for a crowd of people ;
while that of a private apartment will be wide enough if it
permit one person to pass without being incommoded. Ex-
perience has determined that the least width allowable for
this is 2 feet 8 inches ; although doors leading to inferior
and unimportant rooms may, if circumstances require it, be
as narrow as 2 feet 6 inches ; and doors for closets, where an
entrance is seldom required, may be but 2 feet wide. The
width of the principal door to a public building may be
from 6 to 12 feet, according to the size of the building; and
the width of doors for a dwelling may be from 2 feet 8
inches to 3 feet 6 inches. If the importance of an apart-
ment in a dwelling be such as to require a door of greater
width than 3 feet 6 inches, the opening should be closed
with two doors, or a door in two folds ; generally, in such
cases, where the opening is from 5 to 8 feet, folding or slid-
ing doors are adopted. As to*he height of a door, it should
in no case be less than about 6 feet 3 inches ; and generally
not less than 6 feet 8 inches.
299. — The Proportion between Width and Height: of
single doors, for a dwelling, should be as 2 is to 5 ; and, for
entrance-doors to public buildings, as I is to 2. If the
width is given and the height required of a door for a
3i6
DOORS AND WINDOWS.
dwelling, multiply the width by 5, and divide the product
by 2 ; but if the height is given and the width required,
divide by 5 and multiply by 2. Where two or more doors
of different widths show in the same room, it is well to pro-
portion the dimensions of the more important by the above
rule, and make the narrower doors of the same height as
the wider ones; as all the doors in a suit of apartments,
except the folding or sliding doors, have the best appear-
ance when of one height. The proportions for folding or
sliding doors should be such that the width may be equal
to |- of the height ; yet this rule needs some qualification ;
for if the width of the opening be greater than one half the
width of the room, there will not be a sufficient space left
FIG. 179.
for opening the doors ; also, the height should be about one
tenth greater than that of the adjacent single doors.
300. — Panefls. — Where doors have but two panels in
width, let the stiles and muntins be each \- of the width ; or,
whatever number of panels there may be, let the united
widths of the stiles and the muntins, or the whole width of
the solid, be equal to -| of tt^e width of the door. Thus : in
a door 35 inches wide, containing two panels in width, the
stiles should be 5 inches wide ; and in a door 3 feet 6 inches
wide, the stiles should be 6 inches. If a door 3 feet 6
inches wide is to have 3 panels in width, the stiles and
muntins should be each 4^- inches wide, each panel being 8
inches. The bottom rail and the lock-rail ought to be each
equal in width to TV of the height of the door ; and the top
TRIMMINGS FOR DOORS. 317
rail, and all others, of the same width as the stiles. The
moulding on the panel should be equal in width to j of the
width of the stile.
301. — TTrimmings. — Fig. 179 shows a method of trimming
doors : a is the door-stud ; b, the lath and plaster ; <:, the
ground ; d, the jamb ; ^, the stop ; /and g, architrave casings ;
and h, the door -stile. It is customary in ordinary work to
form the stop for the door by rebating the jamb. But when
the door is thick and heavy, a better plan is to nail on a
piece as at e in the figure. This piece can be fitted to the
door and put on after the door is hung ; so, should the door
be a trifle winding, this will correct the evil, and the door be
made to shut solid.
Fig. 1 80 is an elevation of a door and trimmings suitable
for the best rooms of a dwelling. (For trimmings generally,
see Sect. V.) The number of panels into which a door
should be divided may be fixed at pleasure ; yet the present
style of finishing requires that the number be as small as a
proper regard for strength will admit. In some of our best
dwellings, doors have been made having only two upright
panels. A few years' experience, however, has proved that
the omission of the lock-rail is at the expense of the strength
and durability of the door ; a four-panel door, therefore, is
the best that can be made.
302. — Hanging l>oor*. — Doors should all be hung so as
to open into the principal rooms ; and, in general, no door
should be hung to open into the hall, or passage. As to the
proper edge of the door on which to affix the hinges, no
general rule can be assigned.
WINDOWS.
303. — Requirement* for Light. — A window should be
of such dimensions, and in such a position, as to admit a
sufficiency of light to that part of the apartment for which
it is designed. No definite rule for the size can well be
given that will answer in all cases ; yet, as an apprpxima-
DOORS AND WINDOWS.
tion, the following has been used for general purposes.
Multiply together the length and the breadth in feet of the
apartment to be lighted, and the product by the height in
FIG. 1 80.
feet; then the square root of this product will show the
required number of square feet of glass.
304. — Winclow-Frame§. — For the size of window-frames,
add 4-J inches to the width of the glass for their width, and
WIDTH OF .INSIDE SHUTTERS. 319
6£ inches to the height of the glass for their height. These
give the dimensions, in the clear, of ordinary frames for 12-
light windows ; the height being taken at the inside edge of
the sill. In a brick wall, the width of the opening is 8
inches more than the width of the glass — 4^ for the stiles of
the sash, and 3^ for hanging stiles — and the height between
the stone sill and lintel is about io| inches more than the
height of the glass, it being varied according to the thick-
ness of the sill of the frame.
305. — Inside Shuiter§. — Inside shutters folding into
boxes require to have the box-shutter about one inch wider
than the flap, in order that the flap may not interfere when
both are folded into the box. The usual margin shown be-
tween the face of the shutter when folded into the box and
the quirk of the stop-bead, or edge of the casing, is half an
inch ; and, in the usual method of letting the whole of the
thickness of the butt hinge into the edge of the box-shutter,
it is necessary to make allowance for the tlirow of the hinge.
This may, in general, be estimated at \ of an inch at each
hinging ; which being added to the margin, the entire width
of the shutters will be i J inches more than the width of the
frame in the clear. Then, to ascertain the width of the box-
shutter, add i-J inches to the width of the frame in the clear,
between the pulley-stiles ; divide this product by 4, and add
half an inch to the quotient, and the last product will be
the required width. For example, suppose the window to
have 3 lights in width, 1 1 inches each. Then, 3 times 1 1 is
33, and 4^ added for the wood of the sash gives 37^ ; 37^
and 1^ is 39, and 39 divided by 4 gives 9! ; to which add
half an inch, and the result will be loj inches, the width
required for the box-shutter.
306. — Proportion: Width and Height. — In disposing
and locating windows in the walls of a building, the rules of
architectural taste require that they be of different heights
in different stories, but generally of the same width. The
windows of the upper stories should all range perpendicu-
larly over those of the first, or principal, story ; and they
320 DOORS AND WINDOWS.
should be disposed so as to exhibit a balance of parts
throughout the front of the building. To aid in this it is
always proper to place the front door in the middle of the
front of the building ; and, where the size of the house will
admit of it, this plan should be adopted. (See the latter
part of Art. 50.) The proportion that the height should
bear to the width may be, in accordance with general usage,
as follows :
of the width.
The height
of basement windows, i -J
« i<
principal-story "
2*
« n
second -story "
'1
« u
third-story
If
« «
fourth-story "
I|
« «
attic-story "
th(
the same as the width.
But, in determining the height of the windows for the
several stories, it is necessary to take into consideration the
.height of the story in which the window is to be placed.
For, in addition to the height from the floor, which is gen-
erally required to be from 28 to 30 inches, room is wanted
above the head of the window for the window-trimming
and the cornice of the room, besides some respectable space
which there ought to be between these.
307. — Circular Heads. — Doors and windows usually ter-
minate in a horizontal line at top. These require no special
directions for their trimmings. But circular-headed doors
and windows are more difficult of execution, and require
some attention. If the jambs of a door or window be placed
at right angles to the face of the wall, the edges of the soffit,
or surface of the head, would be straight, and its length be
found by getting the stretch-out of the circle (Art. 524) ;
but when the jambs are placed obliquely to the face of the
Avail, occasioned by the demand for light in an oblique
direction, the form of the soffit will be obtained by the fol-
lowing article ; as also when the face of the wall is circular,
as shown in the succeeding figure.
OBLIQUE SOFFITS OF WINDOWS.
321
308. — Form of Soffit for Circular Window-Heads. —
When the light is received in an oblique direction, let abed
(Fig. 181) be the ground-plan of a given window, and efa a
vertical section taken at right angles to the face of the jambs.
FIG. 181.
From a, through e, draw ag at righ't angles to ab\ obtain
the stretch-out of efa, and make eg equal to it; divide eg-
and efa each into a like number of equal parts, and drop
perpendiculars from the points of division in each ; from
the points of intersection, i, 2, 3, etc., in the line ad,
FIG. 182.
draw horizontal lines to meet corresponding perpendicu-
lars from eg\ then those points of intersection will give the
curve line dg, which will be the on*3 required for the edge
of the soffit. The other edge, cht is found in the same
manner.
322 DOORS AND WINDOWS.
For the form of the soffit for circular window-heads,
when the face of the wall is curved, let abed (Fig. 182) be
the ground-plan of a given window, and e f a a vertical sec-
tion of the head taken at right angles to the face of the
jambs. Proceed as in the foregoing article to obtain the
line dg\ then that will be the curve required for the edge of
the soffit, the other edge being found in the same manner.
If the given vertical section be taken in a line with the
face of the wall, instead of at right angles to the face of the
jambs, place it upon the line cb (Fig. 181), and, having drawn
ordinates at right angles to cb, transfer them to efa ; in this
way a section at right angles to the jambs can be obtained.
SECTION V.— MOULDINGS AND CORNICES.
MOULDINGS.
3O9, — mouldings: are so called because they are of the
same determinate shape throughout their length, as though
the whole had been cast in the same mould or form. The
regular mouldings, as found in remains of classic architec-
ture, are eight in number, and are known by the following
names :
FIG 183. Annulet, band, cincture, fillet, listel or square.
FIG. 184. ^ Astragal or bead.
Torus or tore.
FIG. 185.
FIG 186 Scotia, trochilus or mouth.
Ovolo, quarter-round or echinus.
FIG. 187.
Cavetto, cove or hollow.
FIG. 188.
324 MOULDINGS AND CORNICES.
Cymatium, or cyma-recta.
FIG. 189.
Ogee.
Inverted cymatium, or cyma-reversa.
FIG. 190.
Some of the terms are derived thus : Fillet, from the French
word fil, thread. Astragal, from astragalos, a bone of the
heel — or the curvature of the heel. Bead, because this
moulding, when properly carved, resembles a string of beads.
Torus, or tore, the Greek for rope, which it resembles when
on the base of a column. Scotia, from skotia, darkness, be-
cause of the strong shadow which its depth produces, and
which is increased by the projection of the torus above it.
Ovolo, from ovum, an egg, which this member resembles,
when carved, as in the Ionic capital. Cavetto, from cavus,
hollow. Cymatium, from kumaton, a wave.
310. — Characteristics of Mouldings. — Neither of these
mouldings is peculiar to any one of the orders of architect-
ure ; and although each has its appropriate use, yet it is by
no means confined to any certain position in an assemblage
of mouldings. The use of the fillet is to bind the parts, as
also that of- the astragal and torus, which resemble ropes.
The ovolo and cyma-reversa are strong at their upper ex-
tremities, and are therefore used to support projecting parts
above them. The cyma-recta and cavetto, being weak at
their upper extremities, are not used as supporters, but are
placed uppermost to cover and shelter the other parts. The
scotia is introduced in the base of a column to separate the
upper and lower torus, and to produce a pleasing variety
and relief. The form of the bead and that of the torus is the
same ; the reasons for giving distinct names to them are
that the torus, in every order, is always considerably larger
than the bead, and is placed among the base mouldings,
GRECIAN MOULDINGS.
325
whereas the bead is never placed there, but on the capital or
entablature ; the torus, also, is seldom carved, whereas the
bead is ; and while the torus among the Greeks is frequently
elliptical in its form, the bead retains its circular shape. While
the scotia is the reverse of the torus, the cavetto is the re-
verse of the ovolo, and the cyma-recta and cyma-reversa are
combinations of the ovolo and cavetto.
FIG. 191.
The curves of mouldings, in Roman architecture, were
most generally composed of parts of circles ; while those of
the Greeks were almost always elliptical, or of some one of
the conic sections, but rarely circular, except in the case of
the bead, which was always, among both Greeks and Ro-
mans, of the form of a semicircle. Sections of the cone af-
ford a greater variety of forms than those of the sphere ; and
perhaps this is one reason why the Grecian architecture so
326
MOULDINGS AND CORNICES.
much excels the Roman. The quick turnings of the ovolo
and cyma-reversa, in particular, when exposed to a bright
sun, cause those narrow, well-defined streaks of light which
give life and splendor to the whole.
311. — A Profile: is an assemblage of essential parts and
mouldings. That profile produces the happiest effect which
FIG, 192.
is composed of but few members, varied in form and size,
and arranged so that the plane and the curved surfaces suc-
ceed each other alternately.
312. — The Grecian Torus and Scotia. — Join the extremi-
ties a and b (Fig. 191), and from /, the given projection of
the moulding, draw/0 at right angles to the fillets ; from b
FJG. 194.
FIG. 195.
draw bh at right angles to a b ; bisect a b in c ; join / and c,
and upon c, with the radius cf, describe the arc/^, cutting
bh'mh', through c draw de parallel with the fillets; make
dc and ce each equal to b //; then de and a b will be conju-
gate diameters of the required ellipse. To describe the
curve by intersection of lines, proceed as directed at Art.
THE GRECIAN ECHINUS.
551 and note ; by a trammel, see Art. 549; and to find the
foci, in order to describe it with a string, see Art. 548.
313. — The Grecian Echinus. — Figs. 192 to 199 exhibit, va-
riously modified, the Grecian ovolo, or echinus. Figs. 192 to
196 are elliptical, a b and b c being given tangents to the curve ;
parallel to which the semi-conjugate diameters, ad and dc,
IMG. 196.
FIG. 197.
are drawn. In Figs. 192 and 193 the lines a d and dc are semi-
axes, the tangents, ab and be, being at right angles to each
other. To draw the curve, see Art. 551. In Fig. 197 the
curve is parabolical, and is drawn according to Art. 560. In
Figs. 198 and 199 the curve is hyperbolical, being described
according to Art. 561. The length of the transverse ax's, a by
FIG. i
FIG. 199.
being taken at pleasure in order to flatten the curve, a b
should be made short in proportion to ac.
314. — The Grecian Cavetto. — In order to describe this,
Figs. 200 and 201, having the height and projection given,
see Art. 551.
315. — The Grecian Cynia-Rccta. — When the projection
is more than the height, as at Fig. 202, make a b equal to the
328
MOULDINGS AND CORNICES.
height, and divide abed into four equal parallelograms ; then
proceed as directed in note to Art. 551. When the projec-
tion is less than the height, draw da (Fig. 203) at right angles
FIG. 201.
FIG. 200.
to ab\ complete the rectangle, abed; divide this into four
equal rectangles, and proceed according to Art. 551.
316. — The Grecian Cyma-Reversa. — When the projection
FIG. 203.
is more than the height, as at Fig. 204, proceed as directed
for the last figure ; the curve being the same as that, the
position only being changed. When the projection is less
FIG. 204.
FIG. 205.
than the height, draw a d (Fig. 205) at right angles to the
fillet ; make a d equal to the projection of the moulding ; then
proceed as directed for Fig. 202.
FORMS OF ROMAN MOULDINGS.
329
317. — Roman Mouldings : are composed pf parts of circles,
and have, therefore, less beauty of form than the Grecian.
The bead and torus are of the form of the semicircle, and the
scotia, also, in some instances ; but the latter is often composed
of two quadrants, having different radii, as at Figs. 206 and
207, which resemble the elliptical curve. The ovolo and ca-
FIG. 206.
FIG. 207.
vetto are generally a quadrant, but often less. When they are
less, as at Fig. 210, the centre is found thus : join the extrem-
ities, a and b, and bisect a b in c ; from c, and at right angles
to a b, draw c d, cutting a level line drawn from a in d ; then d
will be the centre. This moulding projects less than its
height. When the projection is more than the height, as at
Fig. 212, extend the line from c until it cuts a perpendicular
FIG. 208.
FIG. 209.
drawn from a, as at d\ and that will be the centre of the
curve. In a similar manner, the centres are found for the
mouldings at Figs. 2QJ, 211, 213, 216, 217, 218, and 219. The
centres for the curves at Figs. 220 and 221 are found thus:
bisect the line a b at c ; upon a, c and b successively, with a c
or cb for radius, describe arcs intersecting at d and d\ then
those intersections will be the centres.
330
MOULDINGS AND CORNICES.
FIG. 210.
FIG. 211.
FIG. 212.
FIG. 213.
FIG. 214.
FIG. 215
FIG. 216.
FIG. 217.
FORMS OF MODERN MOULDINGS.
331
3(8. — Modern Mouldings: are represented in Figs. 222
to 229. They have been quite extensively and successfully
used in inside finishing. Fig. 222 is appropriate for a bed-
moulding under a low projecting shelf, and is frequently
used under mantel-shelves. The tangent i h is found thus :
bisect the line ab at c, and be at d-, from d draw de at
right angles to eb\ from b draw bf parallel to ed\ upon b,
FIG. 218.
FIG. 219.
with b d for radius, describe the arc d f\ divide this arc
into 7 equal parts, and set one of the parts from s, the limit
of the projection, to o ; make o h equal to o e ; from h, through
c, draw the tangent ki\ divide b h, hc,ci, and ia each into
a like number of equal parts, and draw the intersecting lines
as directed at Art. 521. If a bolder form is desired, draw
the tangent, i h, nearer horizontal, and describe an elliptic
FIG. 220.
FIG. 221.
curve as shown in Figs. 191 and 224. Fig. 223 is much used
on base, or skirting, of rooms, and in deep panelling. The
curve is found in the same manner as that of Fig. 222. In
this case, however, where the moulding has so little projec-
tion in comparison with its height, the point e being found
as in the last figure, h s may be made equal to s e, instead of
o e as in the last figure. Fig. 224 is appropriate for a crown
332
MOULDINGS AND CORNICES.
FIG. 223.
FIG. 224.
PLAIN MOULDINGS.
333
moulding of a cornice. In this figure the height and pro-
jection are given; the direction of the diameter, ab, drawn
FIG. 225.
FIG. 226.
through the middle of the diagonal, ef, is taken at pleasure ;
and dc is parallel to ae. To find the length of dc, draw b h
FIG. 227.
FIG. 228.
FIG. 229.
at right angles toab; upon 0, with of for radius, describe
the arc, ///, cutting bh in h; then make o c and od each
334
MOULDINGS AND CORNICES.
equal to bh.* To draw the curve, see note to Art. 551. Figs.
22$ to 229 are peculiarly distinct from ancient mouldings,
being composed principally of straight lines ; the few curves
they possess are quite short and quick.
Figs. 230 and 231 are designs for antae caps. The di-
ameter of the antas is divided into 20 equal parts, and the
height and projection of the members are regulated in ac-
cordance with those parts, as denoted under H and P, height
and projection. The projection is measured from the mid-
dle of the antse. These will be found appropriate for por-
ticos, doorways, mantelpieces, door and window trimmings,
H.P.
n.
15
8l'l4f!
»*"*i
910J
10
Hi 15
.
^|-H
HH~
910J
FIG. 230.
FIG. 231.
etc. The height of the antae for mantelpieces should be
from 5 to 6 diameters, having an entablature of from 2 to
2J- diameters. This is a good proportion, it being similar to
the Doric order. But for a portico these proportions are
* The manner of ascertaining the length of the conjugate diameter, dc, in
this figure, and also in Figs. 191, 241, and 242 is' new, and is important in this
application. It is founded upon well-known mathematical principles, viz.: All
the parallelograms that may be circumscribed about an ellipsis are equal to
one another, and consequently any one is equal to the rectangle of the two
axes. And again : The sum of the squares of every pair of conjugate diame-
ters is equal to the sum of the squares of the two axes.
EAVE CORNICES.
335
much too heavy : ah antse 1 5 diameters high and an entab-
lature of 3 diameters will have a better appearance.
CORNICES.
319. — Idesigns for Cornice§. — Figs. 232 to 240 are designs
for eave cornices, and Figs. 241 and 242 are for stucco cor-
nices for the inside finish of rooms. In some of these the
projection of the uppermost member from the facia is
divided into twenty equal parts, and the various members
FIG. 232.
are proportioned according to those parts, as figured under
//and P.
320.— Eave Cornices Proportioned to Height of Build-
ing.— Draw the line ac (Fig. 243), and make be and ba each
equal to 36 inches ; from b draw bdvk right angles to ac,
and equal in length to f of a c ; bisect b d in ^, and from a,
through £>, draw af\ upon a, with ac for radius, describe the
arc cf, and upon e, with ef-iwc radius, describe the arc/W;
divide the curve dfc, into 7 equal parts, as at IO, 20, 30,
etc., and from these points of division draw lines to be
336
MOULDINGS AND CORNICES.
J J
FIG. 233.
DDOIfi
JLJLILJLILIULILJLILJU
JUUUUULI
FIG. 234.
EXAMPLES OF CORNICES.
337
FIG. 235.
FIG. 236.
338
MOULDINGS AND CORNICES.
FIG. 237.
H. P.
10J
FIG. 238.
VARIOUS DESIGNS OF CORNICES.
339
H. P.
17
716
3*31
FIG. 239.
H.P.
FIG. 240.
340
MOULDINGS AND CORNICES
H. P.
FIG. 241.
H. P.
FIG. 242.
PROPORTION OF CORNICES.
341
parallel to db; then the distance b i is the projection of a
cornice for a building 10 feet high ; b 2, the projection at 20
feet high ; b 3, the projection at 30 feet, etc. If the projec-
tion of a cornice for a building 34 feet high is required,
divide the arc between 30 and 40 into 10 equal parts, and
V
a b i 2 3 4 c
FIG. 243.
from the fourth point from 30 draw a line to the base, b c,
parallel with bd\ then the distance o/ the point at which
that line cuts the base from b will be the projection re-
quired. So proceed for a cornice of any height within 70
feet. The above is based on the supposition that 36 inches
FIG. 244.
is the proper projection for a cornice 70 feet high. This,
for general purposes, will be found correct ; still, the length
of the line be may be varied to suit the judgment of those
who think differently.
Having obtained the projection of a cornice, divide it
into 20 equal parts, and apportion the several members
342
MOULDINGS AND CORNICES.
according to its destination — as is shown at Figs. 238, 239,
and 240.
32 L — Cornice Proportioned to a given Cornice. — Let
the cornice at Fig. 244 be the given cornice. Upon any
point in the lowest line of the lowest member, as at #, with the
height of the required cornice for radius, describe an intersect-
ing arc across the uppermost line, as at b ; join a and b ;
then b \ will be the perpendicular height of the upper fillet
for the proposed cornice, I 2 the height of the crown mould-
ing— and so of all the members requiring to be enlarged to
the sizes indicated on this line. For the projection of the
\
FIG. 245.
proposed cornice, draw a d at right angles to a b, and c d at
right angles to b c ; parallel with c d draw lines from each
projection of the given cornice to the line ad-, then e d \v\\\
be the required projection for the proposed cornice, and the
perpendicular lines falling upon c d will indicate the proper
projection for the members.
To proportion a cornice according to a larger given cor-
nice, let A (Fig. 245) be the given cornice. Extend a o to b,
and draw c d at right angles to a b ; extend the horizontal
lines of the cornice, A, until they touch o d\ place the height
of the proposed cornice from o to e, and join /"and e\ upon
o, with the projection of the given cornice, o a, for radius,
TO FIND THE ANGLE BRACKET.
describe the quadrant ad\ from d draw db parallel tofe;
upon ot with o b for radius, describe the quadrant b c ; then
o c will be the proper projection for the proposed cornice.
Join a and c ; draw lines from the projection of the different
members of the given cornice to ao parallel to od\ from
these divisions on the line ao draw lines to the line oc
parallel to a c ; from the divisions on the line of draw lines
to the line o e parallel to the line fe\ then the divisions on
the lines o e and o c will indicate the proper height and pro-
jection for the different members of the proposed cornice.
In this process, we have assumed the height, o e, of the pro-
posed cornice to be given; but if the projection, o c, alone
FIG. 247.
be given, we can obtain the same result by a different pro-
cess. Thus: upon o, with oc for radius, describe the quad-
rant c b ; upon <?, with o a for radius, describe the quadrant
ad\ join d and b ; from /draw fe parallel to.db\ then oe
will be the proper height for the proposed cornice, and the
height and projection of the different members can be
obtained by the above directions. By this problem, a cor-
nice can be proportioned according to a smaller given one
as well as to a larger ; but the method described in the pre-
vious article is much more simple for that purpose.
322.— Angle Bracket in a Built Cornice.— Let A (Fig.
246) be the wall of the building, and B the given bracket,
344
OULDINGS AND CORNICES.
which, for the present purpose, is turned down horizontally.
The angle-bracket, C, is obtained thus : through the ex-
tremity, #, and parallel with the wall, fd, draw the line a b ;
make^r equal af, and through c draw cb parallel with ed\
join d and £, and from the several angular points in B draw
ordinates to cut db in i, 2, and 3 ; at those points erect lines
perpendicular to db', from h draw kg parallel to fa-, take
FIG. 248.
the ordinates, i 0, 2 0, etc., at B, and transfer them to C, and
the angle-bracket, C, will be denned. In the same manner,
the angle-bracket for an internal cornice, or the angle-rib of
a coved ceiling, or of groins, as at Fig. 247, can be found.
323.— Raking mouldings matched with Level Returns—
Let A (Fig. 248) be the given moulding, and A b the rake of
CROWN MOULDING ON THE RAKE. 345
the roof. Divide the curve of the given moulding into any
number of parts, equal or unequal, as at i, 2, and 3 ; from
these points draw horizontal lines to a perpendicular erected
from c ; at any convenient place on the rake, as at B, draw
a c at right angles to A b ; also from b draw the horizontal
line b a ; place the thickness, da, of the moulding at A from
b to a, and from a draw the perpendicular line a e ; from
the points i, 2, 3, at A, draw lines to C parallel to A b ;
make a i, a 2, and a 3, at B, and at C, equal to ^ui, etc., at A ;
through the points, i, 2, and 3, at B, trace the curve — this
will be the proper form for the raking moulding, From i,
2, and 3, at C, drop perpendiculars to the corresponding
ordinates from i, 2, and 3, at A ; through the points of inter-
section, trace the curve — this will be the proper form for the
return at the top.
'
PART II.
SECTION VI.— GEOMETRY.
324. — Mathematics Essential. — In this and the following
Sections, which will constitute Part II., there are treated of
certain matters which may be considered as elementary.
They are all very necessary to be understood and acquired
by the builder, and are here compactly presented in a shape
which, it is believed, will aid him in his studies, and at the
same .time prove to be a great convenience as a matter of
reference.
The many geometrical forms which enter into the
composition of a building suggest a knowledge of .Elemen-
tary Geometry as essential to an intelligent comprehension
of its plan and purpose. One of the prime requisites of a
building is stability, a quality which depends upon a proper
distribution of the material of which the building is con-
structed ; hence a knowledge of the laws of pressure and
the strength of materials is essential ; and as these are based
upon the laws of proportion and are expressed more con-
cisely in algebraic language, a knowledge of Proportion and
of Algebra are likewise indispensable to a comprehensive
understanding of the subject. There will be found in this
work, however, only so much of these parts of mathematics
as have been deemed of the most obvious utility in the
Science of Building. For a more exhaustive treatment of
the subjects named, the reader is referred to the many able
works, readily accessible, which make these subjects their
specialties.
325.— Elementary Geometry.— In all reasoning defini-
tions are necessary, in order to insure in the minds of the
348 GEOMETRY.
proponent and respondent identity of ideas. A corollary is
an inference deduced from a previous course of reasoning.
An axiom is a proposition evident at first sight. In the fol-
lowing demonstrations there are many axioms taken for
granted (such as, things equal to the same thing are equal to
one another, etc.) ; these it was thought not necessary to
introduce in form.
326. — Definition. — If a straight line, as A B (Fig. 249),
stand upon another straight line, as CD, so that the two
angles made at the point B are equal — A B C to A B D (Art.
499, obtuse angle] — then each of the two angles is called a
right angle.
327. — Definition.— The circumference of every circle
is supposed to be divided into 360 equal parts, called
degrees ; hence a semicircle contains 180 degrees, a quad-
rant 90, etc.
C B D
FIG. 250.
328. — Definition. — The measure of an angle is the num-
ber of degrees contained between its two sides, using the
angular point as a centre upon which to describe the arc.
Thus the arc C E (Fig. 250) is the measure of the angle
C B E, E A of the angle E B A, and A D of the angle A B D.
329. — Corollary. — As the two angles at B (Fig. 249) are
right angles, and as the semicircle, CAD, contains 180 de-
grees (Art. 327), the measure of two right angles, therefore, is
HE
'UNIVERSITY
RI6HT ANGLES AND OBLIQUE ANG^§^
^4, .._.
1 80 degrees ; of one right angle, 90 degrees ; of half a right
angle, 4$ ; of one third of a right angle, 30, etc.
330. — Definition. — In measuring an angle (Art. 328), no
regard is to be had to the length of its sides, but only to the
degree of their inclination. Hence equal angles are such as
have the same degree of inclination, without regard to the
length of their sides.
331. — Axiom.— If two straight lines parallel to one
another, as A B and CD (Fig. 251), stand upon another
straight line, as E F, the angles A B F and CDF are equal,
and the angle A B E is equal to the angle CD E.
332. — Definition. — If a straight line, as A B (Fig. 250),
stand obliquely upon another straight line, as CD, then one
A C
/
B ~ D
FIG. 251.
of the angles, as A B C, is called an obtuse angle, and the
other, as A B D, an acute angle.
333.— Axiom.— The two angles ABDzndABC (Fig. 250)
are together equal to two right angles (Arts. 326, 329) ; also,
the three angles A B D, E B A, and CBE are together
equal to two right angles.
334.. corollary. — Hence all the angles that can be
made upon one side of a line, meeting in a point in that
line, are together equal to two right angles.
335, corollary. — Hence all the angles that can be made
on both sides of a line, at a point in that line, or all the
angles that can be made about a point, are together equal to
four right angles.
350
GEOMETRY.
336. — Proposition. — If to each of two equal angles a
third angle be added, their sums will be equal. Let ABC
and D E F (Fig. 252) be equal angles, and the angle I J K the
one to be added. Make the angles G B A and H E D each
equal to the given angle IJ K\ then the angle G B C will be
equal to the angle HE F; for if ABC and D E F be angles
A~
C
FIG. 252.
of 90 degrees, and IJK 30, then the angles GBC and
HEF will be each equal to 90 and 30 added, viz., 120
degrees.
337. — Proportion. — Triangles that have two of their
sides and the angle contained between them respectively
equal, have also their third sides and the two remaining
FIG. 253.
angles equal ; and consequently one triangle will every way
equal the other. Let ABC (Fig. 253) and DEF be two
given triangles, having the angle at A equal to the angle at
D, the side A B equal to the side D E, and the side A C
equal to the side D F\ then the third side of one, B C, is equal
to the third side of the other, E F\ the angle at B is equal to
the angle at £* and the angle at C is equal to the angle at
EQUAL TRIANGLES IN PARALLELOGRAMS. 351
F. For if one triangle be applied to the other, the three
points B, A, C, coinciding with the three points E, D, F, the
line Bf must coincide with the line EF\ the angle at B
with the angle at E; the angle at C with the angle at F\
and the triangle B A C be every way equal to the triangle
EDF.
338. — Proposition. — The two angles at the base of an
isosceles triangle are equal. Let ABC (Fig. 254) be an
B D C
FIG. 254.
isosceles triangle, of which the sides, A B and A C, are equal.
Bisect the angle (Art. 506) BA C by the line A D. Then, the
\ineJ5A being equal to the line A C, the line A D of the
triangle E being equal to the line A D of the triangle F
(being common to each), the angle BAD being equal to the
angle DA C, — the line B D must, according to Art. 337, be
D D c
FIG. 255.
equal to the line D C, and the angle at B must be equal to
the angle at C.
339. — Proportion. — A diagonal crossing a parallelogram
divides it into two equal triangles. Let C D E F (Fig. 255)
be a given parallelogram, and C F a line crossing it diag-
onally. Then, as E C is equal to F D, and EF to CD, the
angle at E to the angle at D, the triangle A must, according
to Art. 337, be equal to the triangle B.
352
GEOMETRY.
340 — Proposition.— Let J KL M (Fig. 256) be a given
parallelogram, and K L a diagonal. At any distance between
J K and L M draw N P parallel to J K\ through the point
G, the intersection of the lines KL and N P, draw HI
parallel to K M. In every parallelogram thus divided, the
parallelogram A is equal to the parallelogram B. For, ac-
cording to Art. 339, the triangle JKL is equal to the tri-
angle K L M, the triangle C to the triangle D, and E to F\
M
this being the case, take D and F from the triangle K LM,
and C and £ from the triangle JKL, and what remains in
one must be equal to what remains in the other ; therefore,
the parallelogram A is equal to the parallelogram B.
34-1. — Proposition. — Parallelograms standing upon the
same base and between the same parallels are equal. Let.
A BCD and EFCD (Fig. 257) be given parallelograms
FIG. 257.
standing upon the same base, CD, and between the same
parallels, A F and CD. Then A B and E F, being equal to
CD, are equal to one another; BE being added to both
A B and E F, A E equals B F; the line A C being equal to
B D, and A E to B F, and the angle C A E being equal (Art.
331) to the angle D B F, the triangle A E C must be equal
(Art. 337) to the triangle B F D\ these two triangles being
equal, take the same amount, the triangle B E G, from each,
TRIANGLE EQUAL TO QUADRANGLE.
353
and what remains in one, A B G C, must be equal to what
remains in the other, E F D G\ these two quadrangles being
equal, add the same amount, the triangle C G D, to each, and
they must still be equal ; therefore, the parallelogram
A B CD is equal to the parallelogram E F C D.
342. — Corollary. — Hence, if a parallelogram and triangle
stand upon the same base and between the same parallels,
H
D
FIG. 258.
the parallelogram will be equal to double the triangle.
Thus, the parallelogram A D (Fig. 257) is double (Art. 339)
the triangle C E D.
343. proposition.— Let FG H D (Fig. 258) be a given
quadrangle with the diagonal F D. From G draw G E
FIG. 259.
parallel toFD- extend HD to E\ join F and E ; then the
triangle .F^^will be equal in area to the quadrangle
FGHD. For since the triangles FDG and FDE stand
upon the same base, F D, and between the same parallels,
FD and G E, they are therefore equal (Arts. 341, 342); and
since the triangle C is common to both, the remaining tri-
354 GEOMETRY.
angles, A and B, are therefore equal ; then, B being equal to
A, the triangle F E H is equal to the quadrangle F G H D.
344. — Proposition. — If two straight lines cut each other,
as FG and H J (Fig. 259), the vertical, or opposite angles,
A and C, are equal. Thus, FE, standing upon H y, forms
the angles B and C, which together amount (Art. 333) to two
right angles ; in the same manner, the angles A and B form
two right angles ; since the angles A and B are equal to B
and C, take the same amount, the angle B, from each pair,
and what remains of one pair is equal to what remains of
the other ; therefore, the angle A is equal to the angle C.
The same can be proved of the opposite angles B and D.
345. — Propo8ition. — The three angles of any triangle are
equal to two right angles. Let ABC (Fig. 260) be a given
triangle, with its sides extended to F, E and D, and the line
C G drawn parallel to BE. As G C is parallel, to EB, the
angle at //is equal (Art. 331) to the angle at L ; as the lines
FC and BE cut one another at A, the opposite angles at M
and TV" are equal (Art. 334) ; as the angle at N is equal (Art.
331) to the angle at Jy the angle at J is equal to the angle at
M] therefore, the three angles meeting at £7 are equal to the
three.angles of the triangle A B C ; and since the three angles
at C are equal (Art. 333) to two right angles, the three angles
of the triangle ABC must likewise be equal to two right
angles. Any triangle can be subjected to the same proof.
346. — Corollary. — Hence, if one angle of a triangle be a
right angle, the other two angles amount to just one right
angle.
RIGHT ANGLE IN SEMICIRCLE. 355
34-7. — Corollary. — If one angle of a triangle be a 'right
angle and the two remaining angles are equal to one another,
these are each equal to half a right angle.
348. — Corollary. — If any two angles of a triangle amount
to a right angle, the remaining one is a right angle.
349. — Corollary. — If any two angles of a triangle are to-
gether equal to the remaining angle, that remaining angle is
a right angle.
350. — Corollary. — If any two angles of a triangle are each
equal to two thirds of a right angle, the remaining angle is
also equal to two thirds of a right angle.
351. — Corollary. — Hence, the angles of an equilateral
triangle are each equal to two thirds of a right angle.
FIG. 261.
352. — Proposition. — If from the extremities of the di-
ameter of a semicircle two straight lines be drawn to any
point in the circumference, the angle formed by them at that
point will be a right angle. Let ABC (Fig. 261) be a given
semicircle ; and A B and B C lines drawn from the extrem-
ities of the diameter A C to the given point B; the angle
formed at that point by these lines is a right angle. Join
the point B and the centre D ; the lines DA, D B, and D C,
being radii of the same circle, are equal; the angle at A is,
therefore, equal (Art. 338) to the angle at E\ also, the angle
at C is, for the same reason, equal to the angle at F\ the
angle A B C, being equal to the angles at A and C taken to-
gether, must, therefore (Art. 349), be a right angle.
353. — Proportion. — The square on the hypothenuse of a
right-angled triangle is equal to the squares on the two re-
356'
GEOMETRY.
maining sides. Let ABC (Fig. 262) be a given right-angled
triangle, having a square formed on each of its sides ; then
the square BE is equal to the squares NCand GB taken
together. This can be proved by showing- that the parallelo-
gram B L is equal to the square G B ; and that the parallelo-
gram C L is equal to the square H C. The angle C B D is a
right angle, and the angle A B F\s a right angle ; add to each
of these the angle ABC] then the angle F B C will evidently
be equal (Art. 336) to the angle ABD\ the triangle FB C
and the square G B, being both upon the same base, F B, and
between the same parallels, F B and G C, the square G B is
equal (A rt. 342) to twice the triangle F B C '; the triangle
A B D and the parallelogram B L, being both upon the same
FIG. 262.
base, B D, and between the same parallels, B D and A L, the
parallelogram BL is equal to twice the triangle A B D ;
the triangles, FB C and AB D, being equal to one another
(Art. 337), the square G B is equal to the parallelogram B L,
either being equal to twice the triangle F B C or A B D. The
method of proving H C equal to C L is exactly similar — thus
proving the square B E equal to the squares H C and G B,
taken together.
This problem, which is the 4/th of the First Book of
Euclid, is said to have been demonstrated first by Pythago-
ras. It is stated (but the story is of doubtful authority)
that as a thank-offering for its discovery he sacrificed a hun-
dred oxen to the gods. From this circumstance it is some-
times called the Jiecatomb problem. It is of great value in
DIAGONAL OF SQUARE FORMING OCTAGON.
357
the exact sciences, more especially in Mensuration and As-
tronomy, in which many otherwise intricate calculations are
by it made easy of solution.
354. — Proposition. — In an equilateral octagon the semi-
diagonal of a circumscribed square, having its sides coinci-
dent with four of the sides of the octagon, equals the dis-
tance along a side of the square from its corner to the more
remote angle of the octagon occurring on that side of the
square. Let Fig. 263 represent the square referred to ; in
which 0 is the centre of each ; then A O equals A D. To
prove this, it need only be shown that the triangle A O D is
an isosceles triangle having its sides A O and A D equal. The
FIG. 263.
octagon being equilateral, it is also equiangular, therefore
the angles BCO,ECO,ADO, etc., are all equal. Of the
right-angled triangle FEC,FC and FE being equal, the
two angles FECand FCE, are equal (Art. 338), and are
therefore (Art. 347) each equal to half a right angle. In like
manner it may be shown that FA B and FR A are also each
equal to half a right angle. And since FE C and FA B are
equal angles, therefore the lines E C and A B are parallel
(Art. 331,) and hence the angles E CO and A OD are equal.
These being equal, and the angles ECO and A D O being
equal by construction, as before shown, therefore the angles
A OD and A D 0 are equal, and consequently the lines A O
and A D are equal. (Art. 338.)
35$ GEOMETRY.
355. — Proposition. — An angle at the circumference of a
circle is measured by half the arc that subtends it ; that is,
the angle A B C (Fig. 264) is equal to half the angle A D C.
Through the centre D draw the diameter BE. The tri-
angle A B D is an isosceles triangle, A Z?and B D being ra-
dii, and therefore equal ; hence, the two angles at F and G
are equal (Art. 338), and the sum of these two angles is equal
to the angle at H (Art. 345), and therefore one of them, Gt is
equal to the half of H. The angles at H and at G (or A BE)
are both subtended by the arc A E. Now, since the angle
FIG. 264.
at H is measured by the arc A E, which subtends it, there-
fore the half of the angle at H would be measured by the
half of the arc A E ; and since G is equal to the half of H,
therefore G or A BE is measured by the half of the arc A E.
It maybe shown in like manner that the angle E B C is
measured by half the arc E C, and hence it follows that the
angle A B C is measured by half the arc, A C, that sub-
tends it.
356. — Proposition. — In a circle all the inscribed angles,
A, B, and C (Fig. 265), which stand upon the same side of the
EQUAL ANGLES IN CIRCLES.
359
chord DE are equal. For each angle is measured by half
the arc D F E (Art. 355). Hence the angles are all equal.
357. — Corollary. — Equal chords, in the same circle, sub-
tend equal angles.
FIG. 265.
358. — Proposition. — The angle formed by a chord and
tangent is equal to any inscribed angle in the opposite seg-
ment of the circle ; that is, the angle D (Fig. 266) equals the
angle A. Let H F be the chord, and E 6 the tangent ; draw
the diameter y H '; then JH G is a right angle, also J F H \$
3 6o
GEOMETRY.
a right angle. (Art. 352.) The angles A and B together equal
a right angle (Art. 346) ; also the angles B and D together
equal a right angle (equal to the angle J HG) ; therefore, the
sum of A and B equals the sum of B and D. From each of
these two equals, taking the like quantity B, the remainders
A and D are equal. Thus, it is proved for the angle at A ;
it is also true for any other angle ; for, since all other in-
scribed angles on that side of the chord line H F equal the
angle A (Art. 356), therefore the angle formed by a chord
and tangent equals any angle in the opposite segment of the
circle. This being proved for the acute angle D, it is also
true for the obtuse angle EHF-, for, from any point, K (Fig.
267) in the arcHKF, draw lines to 7, F and H ; now, if it can
be proved that the angle EH F equals the angle FK H, the
entire proposition is proved, for the angle FKH equals any
of all the inscribed angles that can be drawn on that side of
the chord. (Art. 356.) To prove, then, that EH F equals
H KF\ the angle EH F equals the sum of the angles A and
B ; also the angle H K F equals the sum of the angles C and
D. The angles B and D, being inscribed angles on the same
chord, J F, are equal. The angles C and A, being right angles
(Art. 352), are likewise equal. Now, since A equals C and B
equals D, therefore the sum of A and B equals the sum of C
and D—OY the angle E H F equals the angle H K F.
359. — Propo§kion. — The areas of parallelograms of
equal altitude are to each other as the bases of the parallelo-
PARALLELOGRAMS PROPORTIONATE TO BASES.
361
grams. In Fig. 268 the areas of the rectangles A B CD and
B E D F are to each other as the bases CD and D F. For,
putting the two bases in form of a fraction and reducing this
fraction to its lowest terms, then the numerator and denomi-
nator of the reduced fraction will be the numbers of equal
parts into which the two bases respectively may be divided.
For example, let the two given bases be 12 and 9 feet respect-
ively, then -ijp- = f , and this gives four parts for the larger
base and three parts for the smaller one. So, in Fig. 268,
divide the base CD into four equal parts, and the base D F
into three equal parts ; then the length of any one of the
parts in CD will equal the length of any one of the parts in
D F. Now, parallel with A C, draw lines from each point of
division to the line A E. These lines will evidently divide
the whole figure into seven equal parts, four of them occupy-
FIG. 268.
ing the area A B C D, and three of them occupying the area
B E D F. Now it is evident that the areas of the two rect-
angles are in proportion as the number of parts respectively
into which the base-lines are divided, or that — .
A B C D \ B E D F \ : C D \ D F.
The areas in this particular case are as 4 to 3. But in gen-
eral the proportion will be as the lengths of the bases.
Thus the proposition is proved in regard to rectangles, but
it has been shown (Art. 341) that all parallelograms of equal
base and altitude are equal. Therefore the proposition is
proved in regard to parallelograms generally, including rect-
angles.
360. — Proposition. — Triangles of equal altitude are to
each other as their bases. It has been shown (Art. 359) that
362 GEOMETRY.
parallelograms of equal altitude are in proportion as their
bases, and it has also been shown (Art. 342) that of a triangle
and parallelogram, when of equal base and altitude, the
parallelogram is equal to double the triangle. Therefore
triangles of equal altitude are to each other as their bases.
361. — Proposition. — Homologous triangles have their
corresponding sides in proportion. Let the line CD (Fig.
269) be drawn parallel with. A B. Then the angles E CD
and E A B are equal (Art. 331), also the angles E D C and
E B A are equal. Therefore the triangles E CD and E A B
are homologous, or have their corresponding angles equal.
FIG. 269.
For, join C to- B^ and A to A then the triangles A C D and
B C D) standing on the same base, C D, and between the
same parallels, CD and A B, are equal in area. To each
of these equals join the common area C D E, and the sums
A DE and B CE will be equal. The triangles CD E and
A D E, having the same altitude, are to each other as
their bases C E and A E (Art. 360), or—
CDE : ADE : : CE : A E.
Also the triangles CDE and B C E, having the same alti-
tude, are to each other as their bases D E and BE, or —
CDE ; BCE : : DE : BE.
CHORDS GIVING EQUAL RECTANGLES.
363
And, since the triangles A D E and B C E are equal, as before
shown, therefore, substituting in the last proportion A D E
for B CE, we have —
CD E : AD E : : D E : BE.
The first two factors here being identical with the first two
in the first proportion above, we have, comparing the two
proportions —
CE : AE : : DE : B E\
or, we have the corresponding sides of one triangle, CD E,
in proportion to the corresponding sides of the other, A BE.
FIG. 270.
362. — Proposition. — Two chords, E F and CD (Fig. 270),
intersecting, the parallelogram or rectangle formed by the
two parts of one is equal to the rectangle formed by the
two parts of the other. That is, the product of C G multi-
plied by G D is equal to the product of E G multiplied by
G F. The triangle A is similar to the triangle B, because it
has corresponding angles. The angle H equals the angle G
(Art. 344); the angle at J equals the angle at K, because
they stand upon the same chord, D F (Art. 356) ; for the same
364 GEOMETRY.
reason the angle M equals the angle L, for each stands upon
the same chord, E C. Therefore, the triangle A having the
same angles as the triangle B, the length of the sides of one
are in like proportion as the length of the sides in the other
(Art. 361). So—
CG : EG : : GF : G D.
Hence, as the product of the means equals the product of
the extremes (Art. 373), E G multiplied by GF is equal to
C G multiplied by G D.
363. — Proposition. — If the sides of a quadrangle are
bisected, and lines drawn joining the points of bisection in
the adjacent sides, these lines will form a parallelogram.
FIG. 271.
Draw the diagonals A B and CD (Fig. 271). It will here be
perceived that the two triangles A E O and A C D are homol-
ogous, having like angles and proportionate sides. Two of
the sides of one triangle lie coincident with the two corres-
ponding sides of the other triangle, therefore the contained
angles between these sides in each triangle are identical.
By construction, these corresponding sides are proportion-
ate; AC being equal to twice A E, and A D being equal to
twice A O ; therefore the remaining sides are proportionate,
CD being equal to twice E O, hence the remaining corres-
ponding angles are equal. Since, then, the angles A E O
and A C D are equal, therefore the line E 0 is parallel with
PARALLELOGRAM IN QUADRANGLE. 365
the diagonal CD — so, likewise, the line MNis parallel to the
same diagonal, CD. If, therefore, these two lines, EO and
MN, are parallel to the same line, CD, they must be parallel
to each other. In the same manner the lines ON and EM
are proved parallel to the diagonal A B, and to each other ;
therefore the inscribed figure ME ON is a parallelogram.
It may be remarked, also, that the parallelogram so formed
will contain just one half the area of the circumscribing
quadrangle.
SECTION VII.— RATIO, OR PROPORTION.
364. — Merchandise. — A carpenter buys 9 pounds of nails
for 45 cents. He afterwards buys 87 pounds at the same
rate. How much did he pay for them ?
An answer to this question is readily found by multiply-
ing the 87 pounds by 45 cents, the price of the 9 pounds,
and dividing the product, 3915, by 9 ; the quotient, 435 cents,
is the answer to the question.
365. — The "Rule of Three." — The process by which
this problem is solved is known as the Rule of Three, or
Proportion.
In cases of this kind there are three quantities given,, to
find a fourth. Previous to working the question it is usual
to make a statement, placing the three given quantities in
such order that the quantity which is of like kind with the
answer shall occupy the second place; the quantity upon
which this depends for its value is put m the first place, and
the remaining quantity, which is of like kind with that in the.
first place, is assigned to the third place.
When thus arranged, the second and third quantities are
multiplied together and the product is divided by the first
quantity ; the quotient, the answer to the question, is a
fourth quantity. These four quantities are related to each
other in this manner, namely : the first is in proportion to
the second as the third is to the fourth ; or, taking the
quantities of the given example, and putting them in a for-
mal statement with the customary marks between them, we
have —
9 : 45 : : 87 : 435,
which is read: 9 is to 45 as 87 is to 435 ; or, 9 is in propor-
tion to 45 as 87 is to 435 ; or, 9 bears the same relation to 45
as 87 does to 435.
EQUALITY OF RATIOS. 367
366. — Couple*: Antecedent, Consequent. — These four
quantities are termed Proportionals, and may be divided into
two couples ; the first and second quantities forming one
couple, and the third and fourth the other couple. Of each
couple the first quantity is termed the antecedent, and the
last the consequent. Thus 9 is an antecedent and 45 its con-
sequent ; so, also, 87 is an antecedent and 435 its consequent.
367. — Equal Couples : an Equation. — These four quan-
tities may be put in form thus :
45. = 435
9 - 87
Each couple is here stated as a fraction : each has its ante-
cedent beneath its consequent, and the two couples are
separated by a sign, two short parallel lines, signifying
equality. This is an equation, and is read thus : 45 divided
by 9 is equal to 435 divided by 87 ; or, as ordinary fractions :
45 ninths are equal to 435 eighty-sevenths.
368. — Equality of Ratios. — Each couple is also termed a
Ratio, and the two the Equality of Ratios. Thus the ratio
—is equal to the ratio ^-. If the division indicated in
9 87
these two ratios be actually performed, the equality between
the two will at once be apparent, for the quotient in each
case is 5. The resolution of each couple into its simplest
form by actual division is shown thus :
f-
These are read : 45 divided by 9 equals 5 ; and 435 divided
by 87 equals 5.
369. — Equali multiplied by Equal§ Give Equals. — If two
equal quantities be each multiplied by a given quantity, the
368 RATIO, OR PROPORTION.
two products will be equal. For example, the fractions f
and f are each equal to ^, and are therefore equal to each
other. If these two equal quantities be each multiplied by
any given number, say, for example, by 4, we shall have 4
times f equals f , and 4 times f equals -1/- ; these products, f
and \2- are each equal to 2, and therefore equal to each
other.
370. — Multiplying an Equation. — The quantity on each
side of the sign = is called a member of the equation. If
each member be multiplied by the same quantity, the
equality of the two members is not thereby disturbed (Art.
369); therefore, if the two members of the equation
45 — -415. (Art. 367) be each multiplied by 87, or be modified
9 °7
thus:
45x87^435x87
9 , 87
in which x, the sign for multiplication, indicates that the
quantities between which it is placed are to be multiplied
together ; this ddition to each member of the equation does
not destroy the equality ; the members are still equal,
though considerably enlarged. The equality may be easily
tested by performing the operations indicated in the equa-
tion. For example : for the first member, we have 45 times
87 equals 3915, and this divided by 9 equals 435. Again, for
the second member we have 435 times 87 equals 37845, and
this divided by 87 equals 435, the same result as that for the
first member. Thus the multiplication has not interfered
with the equality of the members.
371. — Multiplying and Dividing one Member of an
Equation: Cancelling. — If a quantity be multiplied by a
given number, and the product be divided by the same
given number, the quotient will equal the original quantity.
For example : if 8 be multiplied by 3, the product will be 24 ;
then if this product be divided by 3, the quotient will be 8,
the original quantity. Thus the value of a quantity is not
TRANSFERRING FACTORS IN EQUATIONS. 369
changed by multiplying it by a number, provided it be also
divided by the same number.
From this, also, we learn that the value of a quantity
which is required to be multiplied and divided by the same
number will not be changed if the multiplication and divis-
ion be both omitted ; one cancels the other. Therefore the
number 87, appearing in the second member of the equation
in the last article both as a multiplier and a divisor, may be
omitted without destroying the equality of the two mem-
bers. The equation thus treated will be reduced to —
45 x87
This expression is read : the product of 45 times 87 divided
by 9 equals 435. It will be observed that we have here the
four terms of the problem in Art. 365, three of them in the
first member, and the fourth, the answer to the problem, in
the second member.
372. — Transferring a Factor .—Each of the four quan-
tities in the aforesaid equation is termed a factor. Compar-
ing the equation of the last article with that of Art. 43, it
appears that the two are alike excepting that the factor 87
has been transferred from one member of the equation to
the other, and that, whereas it was before a divisor, it has
now become a multiplier. From this we learn that a factor
may be transferred from one member of an equation to the
other, provided that in the transfer its relative position to
the horizontal line above or below it be also changed ; that
is, if, before the transfer, it be below the line, it must be put
above the line in the other member; or, if above the line, it
must be put below, in the other member. For example : in
the equation of the last article let the factor 9 be removed
to the second member of the equation. It stands as a divi-
sor in the first member ; therefore, by the rule, it must appear
as a multiplier after the transfer ; or —
45 x 87 = 9 x 435;
3/O RATIO, OR PROPORTION.
which is read, 45 times 87 equals 9 times 435. By actually
performing the operations here indicated, we find that each
member gives the same product, 3915; thus proving that
the equality of the two members was not interfered with
by the transfer.
373. — Equality of Products: Means and Extremes.— In
Art. 366, the four factors are put in the usual form of four
proportionals. A comparison of these with the four factors
as they appear in the equation in the last article, shows that
the first member contains the second and third of the four
proportionals, and the second member contains the first and
the fourth ; or, the first contains what are termed the
means, and the second, the extremes. From this we learn
that in any set of four proportionals, the product of the
means equals the product of the extremes. As for example,
- = i^ ; so, also, - = i£, an equality of ratios : hence the
four factors, 2, 3, 4, 6, are four proportionals, and may be
put thus :
Extreme, /nean, mean, extreme.
2:31:4:6
and, as above stated, the product of the means (3x4=) 12,
equals the product of the extremes (2x6=) 12.
374-. — Homologous Triangles Proportionate. — The
discussion of the subject of Ratios has thus far been con-
fined to its relations with the mercantile problem of Art.
364. The rules of proportion or the equality of ratios
apply equally to questions other than those of a mercantile
character. They apply alike to all questions in which quan-
tities of any kind are comparable. For example, in geome-
try, lines, surfaces, and solids bear a certain fixed relation to
one another, and are, therefore, fit subjects for the rules of
proportion. It is shown, in Art. 361, that the correspond-
ing sides of homologous triangles are in proportion to one
another. Hence, when, of two similar triangles, two corres-
ponding sides and one other side are given, then by the
equality of ratios the side corresponding to this other side
RATIOS APPLIED TO TRIANGLES. 3/1
may be computed. For example : in two triangles, such as
ECD and EAB (Fig. 269), having their corresponding
angles equal, let the side E C, in the triangle ECD, equal 12
feet, and the corresponding side E A, in the triangle E A B,
equal 16 feet, and the side ED, of triangle ECD, equal 14
feet. Now, having these three sides given, how can we find
the fourth ?, Putting them in proportion, we have, as in
Art. 361—
CE : AE : : DE : BE',
and, substituting for the known sides, their dimensions, we
have —
12 : 16 : : 14 : B E ;
and, by Art. 373—
12 x BE = 16 x 14.
Dividing each member by 12, gives —
Performing the multiplication and division indicated, we
have —
Thus we have the fourth side.equal to i8| feet.
375. _ The Steelyard. — An example of lour proportion-
als may also be found in the relation existing between the
arms of a lever and the weights suspended at their ends. A
familiar example of a lever is seen in the common steelyard
used by merchants in weighing goods. This is a bar, A B,
of steel, arranged as in Fig. 272, with hooks and links, and a
suspended platform to carry R, the article to be weighed ;
and with a weight P, suspended by a link at B, from the bar
A B, along which the weight P is movable.
The entire load is sustained by links attached to the ful-
crum, or point of suspension C. The apparatus is in equi-
librium without R and P. In weighing any article, R, the
3/2 RATIO, OR PROPORTION.
weight P is moved along the bar B C until the weight just
balances the load, or until the bar A B will remain in a hori-
zontal position. If the weight P be too far from the fulcrum
C the end of the bar B will fall, but if it be too near it will
rise.
376. — The L<ever Exemplified l>y the Steelyard. — To
exemplify the principle of the lever, let the bar A B (Fig.
272) be balanced accurately with the scale platform, but
without the weights R and P. Then, placing the article R
upon the platform, move the weight P along the beam until
there is an equilibrium. Suppose the distances A C and B C
are found to be 2 and 40 inches respectively, and suppose
FIG. 272.
the weight P to equal 5 pounds, what at this point will be
the weight of R? By trial we shall find that R = 100 pounds.
Again, if a portion of R be removed, then the weight P
would have to be moved along the bar B C to produce an
equilibrium ; suppose it be moved until its distance from C
be found to be 20 inches, then the weight of R would be
found to be 50 pounds, or —
R = 50 pounds.
Again, suppose a part of the weight taken from R be re-
stored, and the weight P, on being moved to a point re-
quired for equilibrium, be found to measure 30 inches from
C, then we shall find that—
R = 75 pounds.
RATIOS OF THE LEVER. 373
Thus when —
B C = 40, R = loo ; or, -- = 2 • 5 ;
40
BC=2o, R = $o>J or, -=2-5;
showing an equality of ratios ; or, in general, B C is in pro-
portion to R) or —
BC : R.
If, instead of moving P along B C, its position be permanent,
and the weight P be reduced as needed to produce equilib-
rium with the various articles, R, which in turn may be
put upon the scale ; then we shall find that if when the
weight P equals 5 pounds the article R equals 100, and there
is an equilibrium, then when —
P-—— x 5 =4-5, R will equal -^*x 100 =-90;
8 8
P = — x 5 = 4, R will equal — x 100 = 80;
P= -^ x 5 r= 3 • 5, R will equal — x 100 = 70 ;
and so on for other proportions; and in every case we shall
r>
have the ratio - equal 20, thus —
R 90
-75- = '^— = 20 '
P 4-5
R 80
P=T = 2°;
^? 70
7 = 3^ = 2°-
374 RATIO, OR PROPORTION.
Thus we have an equality of ratios in comparing the
weights.
Again, if the weight P and the article R be permanent in
weight, and the distances A C, B C be made to vary, then if
there be an equilibrium when A C is 2 and B C is 40, we
shall find that when —
o O
A C = — - x 2 = I -6 ; B C will equal — x 40 = 32 ,
A C ' = — x 2 = i • 2 ; /? C" will equal — x 40 = 24 ;
AC= ~ x2 = °'8; BC wil1 eclual — x 40 = 16 ;
and so on for other proportions, and in every case we shall
BC
have the ratio -jyr = 20 ; thus —
B C 32
_ - ^ - *}f\ .
- _,-, ™~ --• • — • v/ •
^4 T 1-6
^C~ 0-8 •
producing thus an equality of ratios in comparing the arms
of the lever. From these experiments we have found, in
comparing the article weighed with an arm of the lever, the
constant ratio B C : R, and when comparing the weights
we have found the constant ratio P : R. Again, in com-
paring the arms of the lever, we find the constant ratio
A C : B C. Putting two of these couples in proportion, we
have —
A C : B C : : P : R.
Hence (Art. 373) we have—
PRINCIPLE OF THE LEVER DEMONSTRATED. 375
Dividing both members by A C, we have —
BC*P
~AC—
In a steelyard the short arm, A C, and the weight, or poise,
P, are unvarying ; therefore we have—
or, when ^ „• is constant, we have —
R : B C.
377.— The L,cvcr Principle Bcmon§trated. — The rela-
tion between the weights and their arms of leverage may be
demonstrated as follows : *
FIG. 273. .
Let A B G H, Fig. 273, represent a beam of homogeneous
material, of equal sectional area throughout, and suspended
upon an axle or pin at C, its centre. This beam is evidently
in a state of equilibrium. Of the part of the beam A D G K,
let E be the centre of gravity ; and of the remaining part,
D D K H, let F be the centre of gravity.
If the weight of tne material in A D G K\>t concentrated
at E, its centre of gravity, and the weight of the material in
* The principle upon which this demonstration is based may be found in an
article written by the author and published in the Mathematical Monthly, Cam-
bridge, U. S., for»i8s8, p. 77.
376 RATIO, OR PROPORTION.
DBKH be concentrated in F, its centre of gravity, the
state of equilibrium will not be interfered with. Therefore
let the ball R be equal in weight to the part A D G K, and
the ball P equal to the weight of the part D BKH\ and let
these two balls be connected by the rod E F. Then these
two balls and rod, supported at C, will evidently be in a
state of equilibrium (the rod EF being supposed to be with-
out weight).
Now, it is proposed to show that R is to P as C F is to
C E. This can be proved; for, since R equals the area
ADGK and P equals the area DBKH, therefore R is in
proportion to A D, as P is to D B (Art. 359) ; or, taking the
halves of these lines, R is in proportion to A J as P is to
LB.
Also, J L equals half the length of the beam ; for J D is
the half of A D, and D L is the half of DB; thus these two
parts (JD + DL) equal the half of the two parts (AD + DB)\
or, y L equals the half of A B\ or, we have —
Adding these two equations together, we have —
Now, JD + DL = JL, and AD + DB = AB\ therefore,
Thus we have A M = J L. From each of these equals
take J M, common to both, then the remainders, A J and
ML, will be equal ; therefore, A J = C F.
We have also MB = J L. From each of these equals
take ML, common to both, and the remainders, J M and
L B, will be equal ; therefore, L B = E C. As was above
shown —
RiAy'iiP-iL.9.
TO FIND A FOURTH PROPORTIONAL. 377
Substituting for A J and LB their values, as just found,
we have—
R : CF : : P : EC',
from which we have (Art. 373) —
Px CF= R x E C.
Thus it is demonstrated that the product of one weight into
its arm of leverage, is equal to the product of the other
weight into its arm of leverage : a proposition which is
known <is the law of the lever.
378. — Any One of Four Proportionals may be Found.
— Any three of four proportionals being given, the fourth
may be found ; for either one of the four factors may be
made to stand alone ; thus, taking the equation of the last
article, if we divide both members by CF (Art. 371), we
have —
Px CF'_RxEC
C.F CF '
In the first member C F, in both numerator and denominator,
cancel each other (Art. 371), therefore—
so likewise we may obtain—
Px CF
SECTION VIII.— FRACTIONS.
379. — A Fraction Defined. — As a fracture is a break or
division into parts, so a fraction is literally a piece broken off;
a part of the whole.
The figures which are generally used to express a frac-
tion show what portion of the whole, or of an integer, the
fraction is : for example, let the line A B, (Fig. 274), be divided
into five equal parts, then the line A C, containing three of
those parts, will be three fifths of the whole line A B, and
» 3
may be expressed by the figures 3 and 5, placed thus, — ,
which is known as a fraction and is read, three fifths. The
number 5 below the line deno'tes the number of parts into
which an integer or unit, A B, is supposed to be divided ; it
riii 1 1
A D E C B
FIG. 274.
is therefore called the denominator, and expresses th3 denom-
ination or kind, whether fifths, sixths, ninths, or any number,
into which a unit is supposed to be divided. The number
3 above the line, denoting the number of parts contained in
the fraction, is termed the numerator, and expresses the
number of parts taken, as 2, 3, 4, or any other number.
380. — Graphical Repre§entation of Fractions : Effect
of Multiplication. — In Fig. 275, let the line A B be di-
vided into three equal parts ; the line CD into six equal
parts; the line EF into nine equal parts; the line G H into
twelve equal parts, and the line y A' into fifteen equal parts.
The lines AB, CD, EF,GH, and J K, being all of equal
length.
FRACTIONS ILLUSTRATED.
379
Then the parts of these lines, A L, CM, EN, etc., may
be expressed respectively by the fractions-,^,-, - and --.
369 12 15
In each case the figure below the line, as, 3, 6, 9, 12, or 15,
expresses the number of parts into which the whole is di-
vided, and the figure above the line, as 1,2, 3, 4, or 5, the
L
M
N
D
P
FIG. 275.
number of the parts taken ; and, as the lines A Z, CM, EN,
etc., are all equal to each other, therefore these fractions are
all equal to each other. If the numerator and denominator
of the first fraction be each multiplied by 2, the products
will equal the numerator and denominator of the second
fraction ; thus —
1X2 = 2
3X2 = 6'
so, also,
and
and
I>< 3 ~3.
3x3=9'
i_x 4= 4^
3 x4= 12
ix 5 =_i_
3 x 5 =-15
Thus it is shown that when the numerator and denomi-
nator of a fraction are each multiplied by the same factor,
the product forms a new fraction which is of equal value
with the original.
In like manner we have, |, — , A --, etc., each equal to
o 12 It) 2O
one fourth; and which may be found by multiplying' the
numerator and denominator of - successively by 2, 3, 4, 5, etc.
4
380 FRACTIONS. •
381. — Form of Fraction Changed by Division. — By an
operation the reverse of that in the last article, we may re-
duce several equal fractions to one of equal value. Thus, if
in each we divide the numerator and denominator by the
same number, we reduce it to a fraction of equal value, but
with smaller factors.
For example, taking the fractions of the last article, f , f,
iV xV> let eacn De divided by a number which will divide
both numerator and denominator without a remainder.*
Thus, ^"^2==I 1~"~3 = l
6 + 2*3' 9-^3 = 3"
_4/r-4= l J_^-5 =1
12-4=3' 15-^5 = 3*
As these fractions are shown (Art. 380) to be equal, and
as the operation of dividing each factor by a common num-
ber produces quotients which in each case form the same
fraction, -J-, we therefore conclude that the numerator and
denominator of a fraction may be divided by a common
number without changing the value of the fraction.
382. — Improper Fractions. — The fractions f , ^, ~, etc.,
all fractions which have the numerator larger than the de-
nominator are termed improper fractions. They are not im-
proper arithmetically, but they are so named because it is an
improper use of language to call that &part which is greater
than the whole.
As expressions of this kind, however, are sabject to the
same rules as those which are fractions proper, it is custom-
ary to include them all under the technical term of fractions.
Expressions like these — all expressions in which one number
is separated by a 'horizontal line from another number below
it, or one set of numbers is thus separated from another set
below it — may be called fractions, and are always to be un-
derstood as indicating division, or that the quantity above
the line is to be divided by the quantity below the line.
Division is indicated by this sign -r-, which is read "divided by."
IMPROPER FRACTIONS. 381
Q 17 2A 3x8x4 17x82
Thus, z> — > — * ^ -» > etc., are all fractions, tech-
nically, although each may be greater than unity. And it is
understood in each case that the operation of division is re-
quired. Thus, - = 3, -- — 8, — — = 4. When the divis-
33 % ,
ion cannot be made without a remainder, then the fraction,
by cutting the numerator into two, may be separated into two
parts, one of which may be exactly divided, and the other
will be a fraction proper. Thus, the fraction -~ is equal to
1 — (for 15 + 2 — 17); and since — equals 3, therefore,
17 15 2 22
— =— + - = 3 + - = 3- So, likewise, the fraction
17x82 __ 1394 :=i375+J9_. .J_9_. J_9_
125 125 125 125 125 125'
383. — Reduction of Mixed Numbers to Fractions — By
an operation the reverse ot that in the last article, a given
mixed number (a whole number and fraction) can be put
into the form of an improper fraction.
This is done by multiplying the whole number by the de-
nominator of the fraction, the product being the numerator
of a fraction equal in value to the whole number ; the de-
nominator of this fraction being the same as that of the given
fraction. The numerator of this fraction being added to the
numerator of the given fraction, the sum will be the numera-
tor of the required improper fraction, the denominator of
which is the same as that of the given fraction. For example,
the required numerator for —
2 J, is 2 x 3 + I — 7. So 2-3- = -J.
2j, is 2 X 4 + I = 9. So 2\ = f.
3i is 3 x 5 +2 = 17. So3f = $.
384. — Division Indicated by the Factors put as a Frac-
tion.— Factors placed in the form of a fraction as — , -, — - or
382 FRACTIONS.
— - indicate division (Art. 382) ; the denominator (the fac-
tor below the line) being the divisor, and the numerator
(the factor above the line) the dividend, while the value of
the fraction is the quotient. Thus of the fraction, — = 20,
9 41
41 is the divisor, 820 the dividend, and 20 the quotient.
From this we learn that division may always be indicated
by placing the factors in the form of a fraction, so that the
divisor shall form the denominator and the dividend the nu-
merator.
385. — Addition of Fractions having Like Denomina-
tors — Let it be required to add the fractions - and -. By
referring to Art. 379 we see that ^4 D (Fig. 274), is one of the
five parts into which the whole line A B is divided ; it is,
therefore, — . We also see that D C contains two of the five
2
parts ; it is, therefore, -. We also see that AD + D C ' — A C,
which contains three of the five parts, or A C = — of A B.
12 3
We therefore conclude that — + — = — . In this operation it
is seen that the denominator is not changed, and that the
resultant fraction has for a numerator a number equal to the
sum of the numerators of the fractions which were required
to be added.
By this it is shown that to add fractions we simply take
the sum of the numerators for the new numerator, making the
denominator of the resultant fraction the same as that of the
fractions to be added. For example : What is the sum of the
fractions — , — and - ? Here we have 14-3+4 — 8 for the
numerator, therefore —
999
SUBTRACTION AND ADDITION OF FRACTIONS. 383
386.— Subtraction of Fractions Of Like Denominators.—
Subtraction is the reverse of addition ; therefore, to sub-
tract fractions a reverse operation is required to that had in
the process of addition ; or simply to subtract instead of
adding.
2 ^
For example, if - be required to be su>tracted from —
we have—
UNIVERSITY
5 5 "
By reference to Fig. 274 an exemplification of^tkis-wiit'
•? 2 T
seen where we have A C = — , A E = — , and E C = — , and
we have —
3 _2 =£
5 5 5'
We therefore have this rule for the substraction of frac-
tions : Subtract the less from the greater numerator ; the remain-
der will be the numerator of the required fraction. The denom-
inator to be the same as that of the given fractions.
387.— Dissimilar Denominators Equalized. — The rules
just given for the addition and subtraction of fractions re-
quire that the given fractions have like denominators.
When the denominators are unlike it is required, before add-
ing or substracting, that the fractions be modified so as to
make the denominators equal. For example : Let it be re-
quired to find the sum of - and -. By reference to Fig.
2 6
275, we find that —on line A B is equal to - on line E F.
These being equal, we may therefore substitute — for -.
Then we have —
6 2_ _ 8
9 + 9 " 9
384 FRACTIONS.
Now, it will be seen that the fraction - may be had by mul-
tiplying both numerator and denominator of the given frac-
2 , 2X^ = 6
tion- by 3, for 3 x - = -;
and we have seen (Art. 380) that this operation does not
change the value of the fraction. From this we learn that
the denominators may be made equal by multiplying the smaller
denominator and Its numerator by any number which will effect
such a result.
For example : ^-+-— = — + — = -^-;
27 14 7 21
3 ,3 7 12 4 7 23 7
4+I7+T6 =:T6+7^+7^=^z '76'
In this example the second fraction is changed by multiply-
ing by i j.
388. — Reduction of Fraction§ to tlieir Lowest Terms. —
The process resorted to in the last article to equalize the
denominators, is not always successful. What is needed for
a common denominator is to find the smallest number
which shall be divisible by each of the given denominators.
Before seeking this number, let each given fraction be
reduced to its lowest terms, by dividing each factor by a
common number. For example: — may, by dividing by 5,
be reduced to — , which is its equivalent. So, also, — , by di-
3 2o
viding by 7, is reduced to — , its lowest terms.
389. — Lea§t Common Denominator. — To find the least
common denominator ^V&w the several fractions in the order
of their denominators, increasing toward the right. If the
largest denominator be not divisible by each of the others,
double it ; if the division cannot now be performed, treble
LEAST COMMON DENOMINATOR. 385
it, and so proceed until it is multiplied by some number
which will make it divisible by each of the other denomina-
tors. This number multiplied by the largest denominator will be
the least common denominator. To raise the denominator of
each fraction to this, divide the common denominator by the de-
nominator of one of the fractions, the quotient will be the
number by which that fraction is to be multiplied, both
numerator and denominator, and so proceed with each frac-
tion. For example : What is the sum of the fractions
-, -, — , -g? One of these, — , may be reduced, by divid-
ing by 2, to ^. Therefore, the series is -, -, -|» ~. On trial
o 2 A. \) o
we find that 8, the largest denominator, is divisible by the
first and by the second, but not by the third, therefore the
largest denominator is to be doubled: 2x8= 16. This is
not yet divisible by the third ; therefore 3 x 8 = 24. This
now is divisible by the third as well as by the first and the
second ; 24 is therefore the least common denominator.
Now dividing 24 by 2, the first denominator, the quotient
12 is the factor by which the terms of the first fraction are
to be raised, or, - ~ — -. For the second we have
24-5-4 — 6, and - ^ — . For the third we have 24 -*- 6 =
4x0 = 24
4, and ~X 7 — J an<^ f°r tne fourth, 24-^-8 = 3, and
o x 4 — 24
7 X 3 — -21
^ ~_— . Thus the fractions in their reduced form are :
12 18 20 21 7i 23
I I I . • ... n ^
24 24 24 24 ~~ 24 ~~ 24*
390. — Lea§t Common Denominator Again. — When the
denominators are not divisible by one another, then to ob-
tain a common denominator, it is requisite to multiply to-
gether all of the denominators which will not divide any of the
other denominators. For example : What is the sum of the
fractions -, -, -, and -?
386
FRACTIONS.
In this case the first denominator will divide the last, but
the others are prime to each other. Therefore, for the
common denominator, multiply, together all but the first ;
or —
5x7x9 = 315 the common denominator ;
and —
315 _:_ 3 — 105, common factor for the first fraction ;
315 -=- 5 = 63, common factor for the second fraction ;
315 _i_ 7 •=. 45, common factor for the third; .
315-5-9 = 35, common factor for the fourth.
And, then —
«
i x 105 = 105 t 2 x 63 = 126 3 x 45 = 135 ^ 4 x 35 = 140
1 x 105 = 315 ' "5 x 63 = 3"i~5 ' 7 x 45 - 3^5 ' 9 x 35 = "
105 126 135 140 _ 506
+ 315 "h 3i5 * 3i5 ~~ 3^5
191
39(.— Fraction§ multiplied Graphically.— Let A B CD
(Fig. 276) be a rectangle of equal sides, or A B equal A C
and each equal one foot. Then A B multiplied by A C will
G
C H (
FIG. 276.
equal the area A B CD, or i x i = i square foot. Let the
line E F be parallel with A B, and midway between -A B and
CD. Then A B x A £ equals half the area of A B CD, or
i x J = -J. Again ; let G H be parallel with E C, and mid-
way between E C and FD. Then E G x E C = i x £ equals
the area E G C H, which is equal to a quarter of the area
MULTIPLICATION OF FRACTIONS.
387
A B C D; or % x % = J; which is a quarter of the superficial
area.
The product here obtained is less than either of the
factors producing it. It must be remembered, however,
that while the factors represent lines, the product represents
superficial area. The correctness of the result may be
recognized by an inspection of the diagram.
392. — Fraction§ multiplied Graphically. — In Fig. 277
let A B equal 8 feet and A C equal 5 feet ; then the rect-
G
FIG. 277.
angle A B CD contains 5 x 8 = 40 feet. The interior lines
divide the space included within A B CD into 40 equal
squares of one foot each. Let A E equal 3 feet or - of A C.
Let A G equal 7 feet or ~ of A B. Then the rectangle
3 7 21
E F A G contains — x '— — — , or twenty-one fortieths of the
5 04°
<+ *j
whole area A B CD. Thus, while the factor fractions -- and -^
5 o
represent lines, it is shown that the product fraction — rep-
40
21 .
resents surface. Thus — is a fraction, E FA G, of the whole
40
surface, CDAB.
393. — Rule for Miitiplication of Fraction*, and Exam-
ple.— In the example given in the last article it will be ob-
388 FRACTIONS.
served that the product of the denominators of the two
given fractions equals the area of the whole figure (A B C D\
while the product of the numerators equals the area of the
rectangle (E 'FA G), the sides of which are equal respec-
tively to the given fractions. From this we obtain for the
product of fractions this—
RULE. — Multiply together the denominators for the new de-
nominator, and the numerators for a new numerator.
j ? j
For example: what is the product of — and — ? Here
we have 20x21—420 for the new denominator, and
7 x 13 = 91 for the new numerator; therefore the product
of—
il x_7 = _£L.
21 2O 420 '
or, of a rectangular area divided one way into 20 parts and
the other way into 21 parts, thus containing 420 rectangles,
1 3 7
the product of the two fractions — and - — is equal to 91 of
these rectangles, or — - of the whole.
394. — Fraction§ Divided Graphically. — Division is the
reverse of multiplication ; or, while multiplication requires
the product of two given factors, division requires one of
the factors when the other and the product are given. Or
(referring to Fig. 277) in division we have the area of the
rectangle, E FA G, and one side, E A; given, to find the
other side, A G.
Now it is required to find the number of times E A is
contained in E FA G. By inspection of the figure we per-
ceive the answer to be, A G times ; for E A xAG — EFA G,
2 I
the given area. Or, when E A F G is given as — and E A
as -, we have as the given problem —
ILL.!
40 ' 5*
DIVISION OF FRACTIONS. 389
Since division is the reverse of multiplication, instead of
multiplying we divide the factors, and have—
21 -i- 3 = 7
40 -r- 5 " 8*
Thus, to divide one fraction by another, for the numerator of
the required factor, divide the numerator of t/ie product by the
numerator of the given factor, and for the denominator of the
required factor divide the denominator of the product by the
denominator of the given factor. For example :
10 . 2 5
Divide ^~ by -. Answer, — .
o Q >| /•»
Divide - - by — . Answer, — .
395. — Rule for Division of Fractions. — The rule just
given does not work well when the factors are not commen-
5 2
surable. For example, if it be required to divide — by — we
have by the above rule—
7-9 " 7 '
9.
Producing fractional numerators and denominators for the
resulting fraction, which require modification in order to
reach those composed only of whole numbers. If the nu-
merators, 5 and 7, of this compound fraction be multiplied
by 9 (the denominator of the denominator fraction), or the
compound fraction by 9, we shall have —
5 ><9
390 FRACTIONS.
And, if these be again multiplied by 2 (the denominator of
the numerator fraction), we shall have —
5 X9
2
— X 2 =
7x9 7x9x2
~9~ ~9~
Like figures above and below in each fraction cancel each
other (Art. 371), therefore, the result reduces to —
5 x 9
7x2'
in which we find the factors of the two original fractions.
In one fraction — we have the factors in position as given,
but in the other — they are inverted. The fraction in which.
the factors are inverted is the divisor. Hence, for division
of fractions, we have this —
RULE. — Invert the factors of the divisor, and then, as in
multiplication, multiply the numerators together for the numera-
tor of the required fraction, and the denominators for the de-
nominator of the required fraction.
c 2
Thus, as before, if - is required to be divided by -, we
have —
ix9 „ 45
7x2 14'
And, to divide — by — , we have —
23 x 9 _ 207
47 x 7 329
2£ 8
Again, to divide — by — , we have —
2$ x 9 _ 225 = _25 = _s
45 x 8 "" 360 "40 "8"
CANCELLING IN ALGEBRA. 39!
This last example has two factors, 9 and 45, one of which
measures the other ; also, the first fraction - - is not in its
45
lowest terms; when reduced it is — . The question, there-
i fore, may be stated thus :
5 x 9 1 .
9 x 8 "" 8 '
for the two 9*8 cancel each other.
SECTION IX.— ALGEBRA.
396. — Algebra Defined. — It occurs sometimes that a
student familiar only with computation by numerals is
needlessly puzzled, in approaching the subject of Algebra,
to comprehend how it is possible to multiply letters together,
or to divide them. To remove this difficulty, it may be suf-
ficient for them to learn that their perplexity arises from a
misunderstanding in supposing the letters themselves are
ever multiplied or divided. It is true that in treatises on
the subject it is usual to speak as though these operations
were actually performed upon the letters. It is always un-
derstood, however, that it is not the letters, but the quan-
tities represented by the letters, which are to be multiplied
or divided.
For example, in Art. 361 it is shown, in comparing similar
sides of homologous triangles, that the bases of the two tri-
angles are to each other as the corresponding sides, or,
referring to Fig. 269, we have C E : A E : : D E : B E.
Now, let the two bases C E and A E be represented respec-
tively by a and b, and the two corresponding sides D E and
B E by c and d respectively ; or, for—
CE : AE : : DE : BE,
put —
a : b : : c : d\
and, by Art. 373, we have —
b x c = a x d,
which may be written—
be = ad\
for x, the sign for multiplication, is not needed between let-
ters, as it is between numeral factors. The operation of
APPLICATION OF ALGEBRA. 393
multiplication is always understood when letters are placed
side by side.
Now, here we have an equation in which, as usually read,
we have the product of b and c equal to the product of a
and d. But the meaning is that the product of the quantities
represented by b and c is equal to the product of the quan-
tities represented by a and d, and that this equation is in-
tended to represent the relation subsisting between the four
proportionals, C £, A E, D E, and BE, of Fig. 269. In order
to secure greater conciseness and clearness, the four small
letters are substituted for the four pair of capital letters,
which are used to indicate the lines of the figures referred to.
397. — Example : Application. — It was shown in the last
article that the four letters a, £, c, and d represent the cor-
responding sides of the two triangles of Fig..2§g, and that —
b c = a d.
Now, let each member of this equation be divided by a, then
(Art. 371)-
If now the dimensions of the three sides represented by a,
b, and c are known, and it is required to ascertain from these
the length of the side represented by d, let the three given
dimensions be severally substituted for the letters repre-
senting them. For example, let a = 40 feet ; b = 52 feet,
and c = 45 feet ; then—
be 52 x 45
d = — = — - = -~ = 58
a 40 40
The quantities being here substituted for the letters ; we have
but to perform the arithmetical processes indicated to obtain
the arithmetical value of d. From this example it is seen
that before any practical use can be made of an algebraical
formula in computing dimensions, it is requisite to substitute
numerals for the letters and actually perform arithmetically
such operations as are only indicated by the letters.
394 ALGEBRA.
398. — Algebra Useful in Constructing Rules. — In all
problems to be solved there are certain conditions or quan-
tities given, by means of which an unknown quantity is to
be evolved. For example, in the problem in Art. 397, there
were three certain lines given to find a fourth, based upon
the condition that the four lines were four proportionals.
Now, it has been found that the relation between quantities
and the conditions of a question can better be stated by let-
ters than by numerals ; and it is the office of algebra to
present by letters a concise statement of a question, and by
certain processes of comparison, substitution and elimina-
tion, to condense the statement to its smallest compass, and
at last to present it in a formula or rule, which exhibits the
known quantities on one side as equal to the unknown on
the other side. Here algebra ends, at the completion of the
rule. To use the rule is the office of arithmetic. For, in
using the rule, each quantity in numerals must be substi-
tuted for the letter representing it, and the arithmetical
processes indicated performed, as was done in Art. 397.
399.— Algebraic Rule§ are General. — One advantage
derived from algebra is that the rules made are general
in their application, For example, the rule of Art. 397,
— = d, is applicable to all cases of homologous triangles,
however they may differ in size or shape from those given in
Fig. 269 — and not only this, but it is also applicable in all
cases where four quantities are in proportion so as to con-
stitute four proportionals. For example, the case of the
four proportionals constituting the arms of a lever and the
weights attached (Arts. 375-378). For, taking the rela-
tion as expressed in Art. 377 —
PxCF= RxEC,
we may substitute for C F the letter n, and for E C the letter
m, then m will represent the arm of the lever E C (Fig. 262),
aid n *he arm of the lever F C. Then we have—
SYMBOLS CHOSEN AT PLEASURE. 395
and from this, dividing by n (/Irt. 372), we have —
or, dividing by m, we have —
(in.) -
v '
m
which is a rule for computing the weight of R, when P and
the two arms of leverage, m and n, are known. For example,
let the weight represented by P be 1200 pounds, the length
of the arm m be 4 feet, and that of n be 8 feet, then we have—
Pn 1 200 x 8
R = - - = - — = 2400 pounds.
m 4
Pn
This rule, R =. — -, is precisely like that in Art. 397 —
— — d — in which three quantities are given to find a fourth,
the four constituting a set of four proportionals.
400. — Symbols €ho§en at Pleasure. — The particular
letter assigned to represent a particular quantity is a matter
of no consequence. Any letter at will may be taken ; but
when taken, it must be firmly adhered to to represent that par-
ticular quantity, throughout all the modifications which may
be requisite in condensing the statement into which it enters
into a formula for use. For example, the two rules named in
Art. 399 are precisely alike — three quantities given to find a
fourth — yet they are represented by different letters. In one,
R and P represent the two weights, and m and ft the arms of
leverage at which they act ; while in the other the letters
a, b,c, and ^represent severally the four lines which constitute
two similar sides of two homologous triangles. The two
rules are alike in working, and they might have been con-
stituted with the same letters. And instead of the letters
chosen any others might have been taken, which con-
venience or mere caprice might have dictated. In some
ALGEBRA.
questions it is usual to put the first letters, as a, b, c, etc., to
represent known quantities, and the last letters, as x, y, z,
for the quantities sought. In works on the strength of
materials it is customary to represent weights by capital
letters, as P, R, U, W, etc., and lines or linear dimensions by
the small letters, as b, d, /, for the breadth, depth, and length,
respectively, of a beam. Any other letters may be put to
represent these quantities, although the initial letter of the
word serves to assist the memory in recognizing the partic-
ular dimensions intended.
40 1. — Arithmetical Processe§ Indicated by Sign§. — In
algebra, the four processes of addition, subtraction, multi-
plication, and division, are frequently required ; and when
the required process cannot be actually performed upon the
letters themselves, a certain method has been adopted by
which the process is indicated. For example, in additon,
when it is required to add a to b, the two letters cannot be
intermingled as numerals may be, and their sum presented ;
but the process of addition is simply indicated by placing
between the two letters this sign, +, which is called plus,
meaning added to ; therefore, to add a to b we have —
which is read a plus b, or the sum of a and b. When the
quantities represented b}^ a and b are substituted for them —
and not till tHen — they can be condensed into one sum.
For example, let a equal 4 and b equal 3, then for—
a-\-b
we have —
4+3;
and we may at once write their sum 7, instead of 4 + 3.
So, likewise, in the process of subtraction, one letter can-
not be taken from another letter so as to show how much of
this other letter there will be left as a remainder ; but the
process of subtraction can be indicated by a sign, as this, — ,
which is called minus, less, meaning subtracted from. For
ALGEBRAICAL SIGNS. 397
example, let it be required to subtract b. from a. To do
this we have —
which is read a minus b, and when the values of a and b are
substituted for them, we have, when a equals 4, and b
equals 3—
a-b,
or —
4-3;
and now, instead of 4 — 3, we may put the value of the
which is unity, or i.
The algebraic signs most frequently used are as follows :
+ ,//#.$•, signifies addition, and that the two quantities be-
tween which it stands are to be added together; as
a + b, read a added to b.
— , minus, signifies subtraction, or that of the two quantities
between which it occurs> the latter is to be subtracted
from the former ; as a — b, read a minus b.
X, multiplied by, or the sign of multiplication. It denotes
that the two quantities between which it occurs are to
be multiplied together ; as a x b, read a multiplied by b,
' or a times b. This sign is usually omitted between
symbols or letters, and is then understood, as a b. This
has the same meaning as a x b. It is never omitted
between arithmetical numbers; as 9x5, read nine
times five.
-^, divided by, or the sign of division, and denotes that of the
two quantities between which it occurs, the former is
to be divided by the latter; as a~b, read a divided by
b. Division is also represented thus :
-, in the form of a fraction. This signifies that a is to be
divided by b. When more than one symbol occurs
above or below the line, or both, as -- , it denotes
that the product of the symbols above the line is to be
divided by the product of those below the line.
39^ ALGEBRA.
= , is equal to, or sign of equality, and denotes that the
quantity or quantities on its left are equal to those on
its right ; as a — b = c, read a minus b is equal to cy or
equals c ; or, 9 — 5 = 4, read nine minus five equals
four. This sign, together with the symbols on each
side of it, when spoken of as a whole, is called an
equation.
a1 denotes a squared, or a multiplied by a, or the second
power of a, and
a* denotes a cubed, or a multiplied by a and again multi-
plied by a, or the third power of a. The small figure,
2, 3, or 4, etc., is termed the index or exponent of the
power. It indicates how many times the symbol is to
be taken. Thus, d1 = a a, a3 = a a a, a" — a a a a.
\/ is the radical sign, and denotes that the square root of the
quantity following it is to be extracted, and
I/ denotes that the cube root of the quantity following it is
to be extracted. Thus, 4/9 = 3, and V27 — 3- The
extraction of roots is also denoted by a fractional in-
dex or exponent, thus —
a1/* denotes the square root "of a,
a* denotes the cube root of a,
a* denotes the cube root of the square of a, etc.
402. — Example in Addition and Subtraction : Cancel-
ling.— Let there be some question which requires a state-
ment to represent it, like this —
a-i-d = c — b,
which indicates that if the quantity represented by a be
added to the quantity represented by d, the sum will be
equal to the quantity represented by c, after there has been
subtracted from it the quantity represented by b ; or, as it is
usually read, a plus d equals c minus b ; or the sum of a and
d equals the difference between c and b. For illustration,
take in place of these four letters, in the order they stand,
the numerals 4, 2, 9, 3, and we shall havq by substitution -
a + d — c — b,
4+2—9 — 3, or adding
and subtracting— 6 = 6.
TRANSFERRING SYMBOLS. 399
If it be required to add to each member of the equation
the quantity represented by b, this will not interfere with
the equality of the members. For a + d are equal to c — d,
and if to each of these two equals a com-mon quantity be
added, the sums must be equal ; therefore —
a + d+ b = c — b-^b,
or by numerals —
4 + 2 + 3 = 9-3 + 3,
or —
9 = 9-
It will be observed that the right hand member contains
the quantity — b and + b. This shows that the quantity b is
to be subtracted and then added. Now, if 3 be subtracted
from 9, the remainder will be 6, and then if 3 be added, the
sum will be 9, the original quantity. Thus it is seen that
when in the same member of an equation a symbol appears as a
minus quantity and also as a plus quantity, the two cancel each
other, and may be omitted. Therefore, the expression—
b = c — b + b
becomes —
a + d+ b = c.
403. — Transferring a Symbol to the Opposite member.
— In comparing, in the last article, the first equation with the
last, it will be seen that the same symbols are contained in
each, but differently arranged : that while in the first equa-
tion b appears in the right hand member and with a minus
or negative sign, in the last equation it appears in the left
hand member and with a plus or positive sign. Thus it is
seen that in the operation performed b has been made to
pass from one member to the other, but in its passage it has
been changed. A similar change may be made with another
of the symbols. For example, from the last equation, let d
be subtracted, or this process indicated, thus —
a + d+b — d = c — d.
400 ALGEBRA.
The plus and minus d, in the left hand member cancel each
other, therefore—
a + b = c — d,
or, by numerals —
4+3=9— 2-
Reducing —
7 = 7.
By this we learn that any quantity (connected by + or — )
may be passed from one member of the equation to the other, pro-
vided the sign be changed.
404. — Signs of Symbols to be Changed when they are
to »e Subtracted, — As an example in subtraction, let the
quantities represented by + b — a —f+ c, be taken from the
quantities represented by + a+b — c— f. This may be
written —
(+a + b — c —f) — (+b — a—f+c),
an expression showing that the quantities enclosed within the
second pair of parentheses are to be .subtracted from those
included within the first pair. Let the quantities represent-
ed in the first pair of parentheses for convenience be repre-
sented by A , or, a + b — c —f — A . Now, by the terms of the
problem, we are required to subtract from A. the quantities
enclosed within the second pair of parentheses. To do this
take first the positive quantity, b, and subtract it or indicate
the subtraction, thus —
A-b;
we will then subtract the positive quantity c, or indicate the
subtraction, thus —
A-b-c.
We have yet to subtract — a and — /, two negative quanti-
ties.
The method by which this can be accomplished may be
discovered by considering the requirements of the problem.
The plus quantities b and £, before being subtracted from A,
were required to have the two negative quantities # and /de-
THE OPERATION TESTED. 40 1
ducted from them. It is evident, therefore, that in subtract-
ing b and c, before this deduction was made, too much has
been taken from A, and that the excess taken is equal to the
sum of a and /. To correct the error, therefore, it is neces-
sary to add just the amount of the excess, or to add the sum
of a and/, or annex them by the plus sign, thus —
A — b — c + a+f.
To test the correctness of the operation as here performed,
let numerals be substituted for the symbols ; let a = 2, b = 3,
c =r i,/= £; then the given quantities to be subtracted, —
become —
(+3 -2-i+i),
which reduces to —
(4 - 24) = ij.
Thus the quantity to be substracted equals ij. Applying
the numerals to the above expression —
A — b+ a +f — c
becomes —
A — 3 + 2 + -J — i =A — 4+2% = A — i-J.
A correct result ; it is the same as before. Restoring now
the symbols represented by A> we have for the whole ex-
pression —
which, by cancelling (Art. 403) and by adding like symbols
with like signs, reduces to —
2 a — 2 c.
To test this result, let the quantity which was represented
by A have the proper numerals substituted, thus :
+ 0 + b — c — /,
+ 2 + 3 ~ i -4=5- i* = 3i-
4O2 ALGEBRA. .
The sum of the given quantity required to be subtracted
was before found to amount to i^, therefore —
A -1$
becomes —
And the result by the symbols as above was —
2 a — 2 c,
which becomes —
2X2 — 2X1,
or —
4 — 2 — 2;
a result the same as before, proving the work correct. An
examination of the signs in the above expression, which de-
notes the problem performed, will show that the sign of each
symbol which was required to be subtracted has been
changed in the operation of subtraction. Before subtract-
ing they were —
after subtraction they are —
(-b + a+f-c).
By this result we learn, that to subtract a quantity we have
but to change its sign and annex it to the quantity from
which it was required to be subtracted.
Example : Subtract a — b from c + d. Answer, c + d — a + b.
If numerals be substituted, say a = 7, b = 4, c = 5, and
d=g, then—
c+d becomes 5+9=14,
a-b " ;_4= 3,
c + d — (a — b) — 14 — 3— ii,
So, also, —
c + d — a + b
becomes —
+ — + = ii.
FRACTIONS ADDED AND SUBTRACTED. 403
405. — Algebraic Fraction*: Added and Subtracted. —
When algebraic fractions of like denominators are to be
added or subtracted, the same rules (Arts. 385 and 386) are
to be observed as in the addition or subtraction of numeri-
cal fractions — namely, add or subtract the numerators for a
new numerator, and place beneath the sum or difference the
common denominator.
For example, what is the sum of T> T» T?
bob
For this we have —
Subtract -> from -;. For this we have —
a a
b-c
d '
What is the algebraical sum of —
ben r^
- - - - and - 5?
For these we have—
b + c — n — r
To exemplify this, let b represent 9, c = 8, n = 2, r = 3,
and d— 12.
Then, for the algebraic sum, we have —
0 + 8 — 2 — 3 12
— = — = i.
12 12
Now, taking the positive and negative fractions sep-
arately, we have—
£ .. *- = !?.
12 "h 12 12 '
and —
n2 nJ^-I
12 12 12 "
404 ALGEBRA
Together —
12. -5 _ _£^ =
12 12 ~~ 12 ~~ '
as before.
406. — The Least Common Denominator. — When the
denominators of algebraic fractions differ it is necessary be-
fore addition or subtraction can be performed to harmonize
them, as in the reduction of the denominators of numerical
fractions (Arts. 388-390). For example, add together the
/7i @ y
fractions 7—, -7, — . In these denominators we perceive
oc o ac
that they collectively contain the letters a, b and c, and no
others. It will be requisite, therefore, that each of the frac-
tions be modified so that its denominator shall have these
three factors. To effect this it will be seen that it is neces-
sary to multiply each fraction by that one of these letters
which is lacking in its denominator. Thus, in the first, a is
lacking, therefore (Art. 380) T— —7—. In the second a
and c are lacking, therefore T — r-, and in the third
by,ac-=-abc
r x b — rb .
b is lacking, therefore — , _ ^-^> Placing them now
together we have —
aa + ace±br a e r
t |
a b c. be b a c
The factor a a may be represented thus #2, which means
that a occurs twice, the small figure at the top indicating
the number of times the letter occurs ; a 2 is called a squared,
a a a = a*, and is called a cubed.
In order to show that the above fraction, resulting as the
sum of the three given fractions, is correct, let a = 2, b — 3,
c = 4, e = 5, and r = 6. Then the three given fractions
are —
2 JU 6 =: 1:+ 1 + 1
3x4 3 2x4 6 3 4
FRACTIONS SUBTRACTED. 405
In equalizing these denominators we multiply the second
fraction by 2, and the third by i£, which will give —
5_ x 2 = 10^ 3x^i_4i.
3x2= 6 ' 4 x i£ ~ ' 6 '
then—
1 • 1°_ ,44 ij_4 03i _7_
6+ 6 " 6 ' 6 6 " 12*
Now the sum of the fractions i
s —
22+2 X4X 5 + 3X6
tJi ,
2X3X4
4 + 40+ 1 8 __ 62 _ 14 _ 7
24 24 " 2 24 ~ 2 12 '
the same result as before, thus showing that the reduction
was rightly made.
407. — Algebraic Fractions Subtracted. — To exemplify
the subtraction of fractions, let it be required to find the
algebraic sum of - - -% — j. These denominators all dif-
fer. The fractions, therefore, require to be modified, so
that each denominator shall contain them all. To accom-
plish this, the first fraction will need to be thus treated :
ax df= adf
7x
the second —
_ b_xcf= _ bcf
~ dXCf= :,. Cdf''
the third—
e x c d = c d e
~ fxcd= ~ 7d~f
The sum of these is —
adf— bcf — c de
~Tdf~ '
406 ALGEBRA.
That this is a correct answer, let the result be proved by
figures ; thus, for a put 15 ; b, 2 ; c, 3 ; d, 4; e, 5 ; /, 6. Then'
we shall have —
a b e 15 25
~c~~d~J~- T " 4" 6*
It will be observed that these denominators may be equal-
ized by multiplying the first fraction by 2, and the second
by ij, therefore we have —
j$o _3 _ 5.
6 " 6 6'
To make the required subtraction we are to deduct from 30
(the numerator of the positive fraction), first 3, then 5 ; or,
the sum of the numerators of the negative fractions ; or for
the numerator of the new fraction we have 30 — 8 — 22.
The required result, therefore, is —
-.~,
63
To apply this test to the algebraic sum we have —
a d f — b c f — c d e I5x4x6 + 2x 3x64.3x4x5
" ~ cdf 3x4x6
which by multiplication reduces to—
360 — 36 — 60 _ 264 _ 22_ ri. a
~W '- 72" : : 6 : : 3 :
a result the same as before, proving the work correct. An
other example :
a b c d . e
From ----- take -, — and - ;
n m n m n
or. find the algebraic sum of—
a . b c d_ e_
n m n , m n
DENOMINATORS HARMONIZED. 407
The fractions which have the same denominator may be
grouped together thus :
a c e a — c — c
n n n n
and —
* A ^L b~d
m m m
To harmonize these two denominators, m and n, the first
fraction must be multiplied by m and the last by ;/, or —
m (a — c — e) n (b — d} m (a — c — e) + n (b —d)
m n m n mn
In the polynomial factor within the parentheses (a — c — e) we
have the positive quantity a, from which is to be taken the
two negatives c and e, or their sum is to be taken from #, or
(a — (c + e) ). With this modification we have for the alge-
braic sum of the five given fractions —
m(a — (c +c)) + n (b — d)
mn
To test the accuracy of this result, let the value of the sev-
eral letters respectively be as follows : a = n, b =g, c = 3,
d = 4, e = 5, m = 10, and n — 8. Then the sum is —
10 (11 -(3 + 5)) + 8 (9-4) __ TO __ 7
10 x 8 "80 8*
Now, taking the fractions separately, we have—
<*__£. £_Ii (A,l\ ii_8 1
n n n ~ 8 V8 + 8/ " 8 8 " 8'
^^945
again- — - - — = = — - — == — ;
or, together we have, as the sum of these two results—
8 + lo'
408
ALGEBRA.
To harmonize these denominators we may multiply the first
fraction by 5, and the second by 4, thus :
8x5=4°' io x 4 = 4°'
and then the sum is —
Jl ™ - 35. = 1_.
40 + 40 "" 40 8 '
the same result as before, thus the accuracy of the work is
established.
408. — Graphical Representation of multiplication. —
In Fig. 278, let A BC D, a rectangle, have its sides A B and
A B
FIG. 278.
A C divided into equal parts. Then the area of the figure
will be obtained by multiplying one side by the other, or
putting a for the side A B, and b for the side A C, then the
area will be a x b, or ab. This will be the correct area of
the figure, whatever the length of the sides may be. If, as
shown, the area be divided into 4 x 7 = 28 equal rectangles,
then a would equal 7, and b equal 4, and a b = 7 x 4 = 28, the
area. If A B equal 28 and A C equal 16, then will a — 28,
and b = 16, and a b = 28 x 16 = 448, the area.
409. — Graphical Multiplication : Three Factor§. — Let
A B C D E FG (Fig. 279) represent a rectangular solid which
may be supposed divided into numerous small cubes as
shown. Now, if a be put for the edge A B, b for the edge
A C, and c for the edge CD, then the cubical solidity of the
MULTIPLICATION OF A BINOMIAL.
409
whole figure will be represented by a x b x c = a b c. If the
edge A B measures 6, the edge A C 3, and the edge CD 4,
then abc = 6x3x4 = 72 = the cubic contents of the
figure, or the number of small cubes contained in it.
DL
/G
FIG. 279.
410. — Graphic Representation: Two and Three Fac-
tors— Figs. 278 and 279 serve to illustrate the algebraic ex-
pressions a b and a b c. In the former it is shown that the
multiplication of two lines produces a rectangular surface,
or that if a and b represent lines, then a b may represent a
rectangular surface (Fig. 278) having sides respectively
equal to a and b. And so if a, b, and c represent three sev-
eral lines, then a b c may represent a rectangular solid
279) having edges respectively equal to #, £, and c.
A BE
FIG. 280.
4f|. — Graphical Multiplication of a Binomial. — Let
A B CD (Fig. 280) be a rectangular surface, and BED F an-
other rectangular surface, adjoining the first. The area of
the whole figure is evidently equal to —
(A B + B E) x A C.
410 ALGEBRA.
The area is also equal to —
ABxA C + BExBD:
or, since A C — B D, the area equals
ABxAC + BExAC;
or, if symbols be put to represent the lines; say a for A B,
b for B E, and c for A C, then the two representatives of the
area, as above shown, become : The first —
(a + b} x c = area ;
and the last —
(a x c) + (b x c) = area.
Hence we have —
(a + b} c = a c -\- b c.
This result exemplifies the algebraic multiplication of a bi-
nomial, which is performed thus : Let a + b be multiplied
by c.
The problem is stated thus :
(a + b) c.
To perform the multiplication indicated we Droceed thus :
a + b
c
ac + be
multiplying each of the factors of the multiplicand sepa-
rately and annexing them by the sign for addition. Putting
the two together, or showing the problem and its answer in
an equation, we have —
(a + b) c == a c + b c,
producing the same result, above shown, as derived from
the graphic representation.
412.— Graphical Squaring of a Binomial.— Let EGCJ
(Fig: 281) be a rectangle of equal sides, and within it draw
SQUARING OF A BINOMIAL. 411
the two lines, A H and F D, parallel with the lines of the
rectangle, and at such a distance from them that the sides,
A B and B D, of the rectangle, A B C D, shall be of equal
length. We then have in this figure the three squares,
E GCJ, AB CD, and FGBH, also the two equal rect-
angles, EFA B and BHD J.
Let E F be represented by a and F G by b, then the area
of ABC D will be axa = a* \ the area of FGBH will be
b x b — b 2 ; the area of E F A B will be a x b = a b, and that
A B
•
C 1
5
FIG. 281.
of B H D y will be the same. Putting these areas together
thus—
the sum equals the area of the whole figure — equals the prod-
uct of EG x E £*— -equals the product —
(a + b) x(a + b).
So, therefore, we have —
(a + b) (a + b} = a* + 2a& + &'; (112.)
or, in general, the square of a binomial equals the square of
the first, plus twice the first by the second, plus the square of the
second. This result is obtained graphically. The same re-
sult may be obtained by algebraic multiplication, combining
412 ALGEBRA.
each factor of the multiplier with each factor of the multi-
plicand and adding the products, thus —
a + b
a + b
a b
The same result as above shown by graphical representa-
tion.
413. — Graphical Squaring of the Difference of Two
Factors. — Let the line E C (Fig. 281) be represented by c,
and the line A E and A C as before respectively by b and
a, then —
c — b — a.
From this, squaring both sides, we have —
The area of the square A B C D may be obtained thus :
From the square E G C J take the rectangle E G x E A and
the rectangle F G x D y, minus the square F G B H, or
from c* take the rectangle cb, and the rectangle c b, minus
the square, b a, and the remainder will be the square, a 8 ; or,
in proper form —
In deducting from c* the rectangle cb twice, we have taken
away the small square twice ; therefore, to correct this
error, we have to add the small square, or b*. Then, when
reduced, the expression becomes —
This result is obtained graphically. The result by algebraic
PRODUCT OF THE SUM AND DIFFERENCE.
413
process will now be sought. The square of a quantity may
be obtained by multiplying the quantity by itself, or —
(c - £)2 = -
(US-)
In this process, as before, each factor of the multiplier is
combined with each factor of the multiplicand and the sev-
eral products annexed with their proper signs (Art. 415),
and thus, by algebraic process, a result is obtained precisely
like that obtained graphically. This result is the square of
the difference of c and b ; and since c and b may represent
any quantities whatever, we have this general —
RULE. — The square of the difference of two quantities is
equal to the sum of the squares of the two quantities, minus twice
their product.
FIG. 282.
4(4. Graphical Product of the Sum and Difference
of Two Quantities. — Let the rectangle A B C D (Fig. 282)
have its sides each equal to a. Let the line E F be parallel
with A B and at the distance b from it, also, the line F G
made parallel with B D, and at the distance b from it. Then
the line E F equals a -f /;, and the line E C equals a — b.
Therefore the area of the rectangle E F C G equals n + b,
4H ALGEBRA.
multiplied by a — b. From the figure, for the area of this
rectangle, we have —
ABCD-ABEH+HFDG = RFC G\
or, by substitution of the symbols,
a 2 — a b + b (a — B).
Multiply the last quantity thus—
a-b
b
ab-b* = b(a-b}.
Substituting this in the above we have —
a* — a b + a b — b* = ( a T &) x (a — b).
Two of these like quantities, having contrary signs, cancel
each other and disappear, reducing the expression to this—
The correctness of this result is made manifest by an inspec-
tion of the figure, in which it is seen that the rectangle E FC G
is equal to the square A BCD minus the square BJHF.
For ABEH equals BJDG. Now, if from the square
A B CD we take away A B E H, and place it so as to cover
BJDG, we shall have the rectangle E FC G plus the square
BJHF-, showing that the square A BCD is equal to the
rectangle EFC ' G plus the square B J H F '; or —
a*=(a + b) x (a-b) + b\
The last quantity may be transferred to the first member of
the equation by changing its sign (Art. 403). Therefore — -
+ b} x (a - b\
as was before shown.
MULTIPLICATION — PLUS AND MINUS.
415
The result here obtained is derived from the geometrical
figure, or graphically. * Precisely the same result may be
obtained algebraically ; thus —
a + b
a- b
a* +ab
-ab-b*
(.114.)
Here the two like quantities, having unlike signs, cancel
each other and disappear, leaving as the result only the dif-
ference of the squares.
The result here obtained is general ; hence we have this —
RuL.E. — The product of the sum and difference of two quan-
tities equals the difference of their squares.
^ G 1
E J
F
FIG. 283.
415. — PIu§ and Minus Sign§ in Multiplication. — In pre-
vious articles the signs in multiplication have been given to
products in accordance with this rule, namely : Like signs
give plus ; unlike signs, minus. This rule may be illustrated
graphically, thus : In the rectangular Fig. 283, let it be re-
quired to show the area of the rectangle A G C H, in terms
of the several parts of the whole figure. Thus the area of
A GE J equal ABEF-GBJF*n& the area of EJCH
equals E FCD - J F H D. And the areas of A G E J -r
EJCH equals the area of A G C H. Therefore the sum of
the two former expressions equals A G C H. Thus A B£F—
GBJF+ EFCD — JFHD = AGCH. Let the several
lines now be represented by algebraic symbols ; for example,
416 ALGEBRA.
let AB = EF=a; let GB = ?F= l>; let A E = G J = c ;
and EC — J H = d, and let these symbols be substituted for
the lines they represent, thus ABEF- GBJF+EFCD -
JFHD = AGCH.
ac — be + ad — b d = (a — b) x (c + d).
An inspection of the figure shows this to be a correct
result. It will now be shown that an algebraical multiplica-
tion of. the two binomials, allotting the signs in accordance
with the rule given, will produce a like result. For example —
a-b
c + d
ac — be + ad — b d.
416. — Equality of Squares on Hj potliemise and Sides of
Right-Angled Triangle. — The truth of this proposition has
been proved geometrically in Art. 353. It will now be
shown graphically and proved algebraically.
Let A BCD (Fig. 284) be a rectangle of equal sides, and
BED the right-angled triangle, the squares upon the sides
of which, it is proposed to consider. Extend the side BE
to F; parallel with BF draw DG, C K, and A L. Parallel
with ED draw A J and L G. These lines produce triangles,
AHB, AC?, ALC, CKD, and C G D, each equal to the
given triangle BED (Art. 337). Now, if from the square
SQUARES ON RIGHT-ANGLED TRIANGLE. 417
A B CD we take Afiffand place it at CD G ; and if we take
BED and place it at A L C we will modify the square
A BCD, so as to produce the figure LGDEHAL, which
is made up of two squares, namely, the square D E FG and
the square ALFH, and these two squares are evidently
equal to the square A B CD. Now, the square D E FG is the
square upon ED, the base of the given right-angled triangle,
and the square A L F H is the square upon A H = BE, the
perpendicular of the given right-angled triangle, while the
square A B C D is the square upon B D, the hypothenuse of
the given right-angled triangle. Thus, graphically, it is shown
that the square upon the hypothenuse of a right-angled triangle
is equal to the sum of the squares upon the remaining two sides.
To show this algebraically, let B E, the perpendicular of
the given right-angled triangle, be represented by a ; E D,
the base, by b, and B D, the hypothenuse, by c. Then it is
required to show that—
Now, since D K == B E = a, therefore, E K = E D -
D K = b — a, and the square E K J H equals (b — #)2, which
(Art. 413) equals
This is the value of the square EKJ H which, with the four
triangles surrounding it, make up the area of the square
A B C D. Placing the triangle A B H of this square outside
of it at CD G, and the triangle B E D at A L C, we have the
four triangles, grouped two and two, and thus forming the
two rectangles C G D K and A L C J. Each of these rect-
angles has its shorter side (A L, C G) equal to BE — a, and
its longer side L C, G D, equal to E D = b ; and the sum of
the two rectangles is ab + ab=2ab. This represents the
area of the two rectangles, which are equal to the four tri-
angles, which, together with the square EKJH, equal the
square ABCD\ or—
ABCD^EKJH+CGDK+ALCJ,
ALGEBRA.
or — c* — (b — a)* + a b + a b, or—
c*= (b - tf)2 + 2ab.
Then, substituting for (b — #)2, its equivalent as above, we
have —
c* = b* — 2ab + a*+ 2ab.
Remove the two like quantities with unlike signs (Art. 402),
and we have —
c*=b*+a*-, (115.)
which was to be proved.
417.— Division the Reverse of Multiplication. — As di-
vision is the reverse of multiplication, so to divide one quan-
tity by another is but to retrace the steps taken in multipli-
cation. If we have the area ab (Fig. 278), and one of the
factors a given to find the other, we have but to remove
from a b the factor a, and write the answer b.
If we have the cubic contents of a solid abc (Fig. 279),
and one of the factors a given to find the area represented
by the other two, we have but to remove a, and write the
others, b c, as the answer.
If there be given the area represented by a (b + c) (see
Art. 41 1), and one of the factors a to find the other, we have
but to remove a and write the answer b + c. Sometimes, how-
ever, a (b + c) is written ab + ac. Then the given factor is
to be removed from each monomial and the answer written
b + c.
If there be given the area represented by a* + 2 ab + b*
to find the factors, then we know by Art. 412 that this area
is that of a square the sides of which measure a+ b, and that
the area is the product of a + b by a + b ; or, that a + b is
the square root of a1 + 2 a b + b*.
If there be given the area a~ •- 2 ab + b~ to find its fac-
tors, then we know by Art. 413 that this area is that of a
square whose sides measure a — b, or that it is the product
of a — b bv a — b, or the square of a — b'
PROCESSES IN DIVISION. 419
If there be given the difference, of the squares of two
quantities, or the area represented by a* — b*y to find its fac-
tors, then we know by Art. 414 that this is the area pro-
duced by the multiplication of a — b by a + b.
4-18. — BH vi si on : Statement of Quotient. — In any case of
division the requirement may be represented as a fraction ;
thus : To divide c + d — /by a — ^ we write the quotient
thus—
c + d-f
a- b
For example, to illustrate by numerals, let a = 7, b = 3,
c = 4, d = 5, and/ = 6. Then the above becomes —
7-3 "4*
4 1 9. — Division ; Reduction. — When each monomial in
either the numerator or denominator contains a common
quantity, that quantity may be removed and placed outside
of parentheses containing the monomials from which it was
taken ; thus, in —
2 ab -\. ^ ac — 8 ad
~T~
we have 2 and a factors common to each monomial of the
numerator. Therefore the expression may be reduced to
2 a (b + 2 c — 4^)
To test this arithmetically we willl put a = 9, b = 7, c — 5,
d = 4, and / = 6. Then for the first expression we have—
2x9x7 + 4x9x5 — 8x9x4
~6~~
which equals —
126 + 1 80 — 288
42O ALGEBRA.
And for the second expression —
2 x 9 (7 + 2 x 5 — 4 x 4)
6
which equals —
18 (17 & 1 6) 18
the same result as before. It will be observed that in this
process of removing all common factors algebra furnishes
the means of performing the work arithmetically with many
less figures. The reduction is greater when the common
factors are found in both numerator and denominator. For
example, in the expression —
$ an + 9 fin — 15 en
12 dn —
we have 3 n a factor common to each monomial in the nu-
merator and denominator ; therefore the expression reduces
to
And now, since 3 n is a factor common to both numerator
and denominator, these cancel each other ; therefore (Art.
371) the expression reduces to—
— 5
To test these reductions arithmetically, let a = 9, b = 8,
c — 4, d — 6, /= 3, and n = .5. Then the first expression
becomes —
3x9x5 _+_9_ x 8 x 5 - 15x4x5
12x6x5 — 18x3x5
which equals —
135 + 360-300^ *95 _ i_.
360 - 270 90 6 '
FORMULA OF THE LEVER. 421
and the second expression becomes —
9 + 3x8 — 5x4
4x6 — 6x3 '
which equals—
9 + 24 — 20 __ ^S __ 1
24—18 6 6*
The same result, but with many less figures.
420. — Proportionals : Analysis. — In the formula of the
lever (Art. 377), P x CF = R x E C. Let n be put for the
arm of leverage 6^and m for E C. Then we have—
Pn = Rm,
from which by division (Art. 372) we have (Art. 399)—
and~
(in.)
Suppose there be a case in which neither R nor P severally
are known, but that their sum is known ; and it is required
from this and the m and n to find R and P. Let—
W = R + P,
then— W- R = P. (See Art. 403.)
The value of P was above found to be—
Since P = R and also equals W - - R, therefore-
422 ALGEBRA.
Transferring R to the opposite member (Art. 403) we hav<
Here R appears as a common factor and may be separated
by division (Art. 419) ; thus—
W= I
By division the factor ( i + — J may be transferred to the
opposite member (Art. 371). Thus we have —
W
by which we find the value of R developed. As an example,
let W = looo pounds, m = 3 feet and n — 7 feet ; then—
1000 _ looo
~- i+l = "ToT '
Multiplying the numerator and denominator by 7, we get—
7 x 1000
=700.
Since — R + P — 1000,
and— R = 700,
then— P — 300.
But a process similar to the above develops an expression
for the value of P, which is —
i + n
~Z ("70
Putting this to the test of figures, we have —
looo looo 3000
r> _ _ — __ — if _ _ — •inn
• i + i - y 10 - 3°°-
NEGATIVE EXPONENTS. 423
4-21. — Raising a Quantity to any Power. — When a
quantity is required to be multiplied by its equal, the prod-
uct is called the square of the quantity. Thus a x a = a*
(Art. 412). If the square be multiplied by the original
quantity the result is a cube ; or, a9 x a = a3 ; or, generally,
for-—
a, a a, a a a, aaaa, aaaaa,
we put—
in which the small number at the upper right-hand cor-
ner indicates the number of times the quantity occurs in
the expression. Thus, if a — 2, then #a = 2 x 2 = 4,
a3 = 4 x 2 = 8, a* = 8 x 2 = 16, a" — 16 x 2 = 32 ; any term
in the series of powers may be found by multiplying the
preceding one by a, or by dividing the succeeding one by a.
Thus a* x a = a*, and - — a\
a
4-22. — Quantities with Negative Exponents. — The series
of powers, by division, may be extended backward. Thus,
a6 a* 3 a* * a* , a1 0 a°
if we divide — = # ; — = a ; - = a ; —= a ', - = a ; - -a ;
a # # a a a
f?lW°; "-=«>, etc.
a a
In this series we have - = a*. But a quantity divided by
its equal gives unity for quotient, or - = i. Therefore, — = i,
and a° — i. This result is remarkable, and holds good re-
gardless of the value of a.
From this and the preceding negative exponents we de-
rive the following :
424 ALGEBRA.
.
a = — = -,
a a
3 a~" I I
a -3=- — — = -, etc.
Showing that # quantity with a negative exponent may have
substituted for it the same quantity with a positive exponent, but
used as a denominator to a fraction having unity for the
numerator.
423. — Addition and Subtraction of Exponential Quan-
tities. — Equal quantities raised to the same power may be
added or subtracted; as, «2 + 2«2= 3#2; but expressions in
which the powers differ cannot be reduced ; thus, a* + a — a*
cannot be condensed.
4-24-. — multiplication of Exponential Quantities. — It
will be observed in Art. 421 that in the series of powers, the
index or exponent increases by unity ; thus, a1, a\ a\ a\ etc. ;
and that this increase is effected by multiplying by the root,
or original quantity. From this we learn that to multiply
two quantities having equal roots we simply add their exponents.
Thus the product of a, a'\ and a3 is a' x a* x a* — a\
The product of a~2, #3, and ab is a~* x a* x a* = a*.
The exponents here, are : — 2 + 3 + 5 — 8 — 2 = 6.
425. — Division of Exponential Quantities. — As division
is the reverse of multiplication, to divide equal quantities
raised to various powers, we need simply to subtract the expo-
nent of the divisor from that of the dividend. Thus, to divide
a" by a9 we have a*"* = a\ That this is correct is manifest ;
for the two factors, a* x a\ in their product, a\ produce the
dividend.
To divide a by a*, we have a^ = a~\ which is equal to -3
EXPLANATION OF LOGARITHMS. 425
(see Art. 422). The same result may be had by stating the
question in the usual form. Thus, to divide a1 by a" we have
— 6, a fraction which is not in the lowest terms, for it may be
put thus, -T— 5 = — , by which it is seen that it has in both its
a a a
numerator and denominator the quantity a\ which cancel
each other (Art. 371). Therefore, -6 = L ; the same result
as before.
426.— Extraction of Radicals.— We have seen that the
square of a is a1 x a1 = a* ; of 2 a3 is 2 a* x 2 a3 = 4 a6 ; in each
case the square is obtained by doubling the exponent.
To obtain the square root the converse follows, namely,
take half of the exponent.
Thus the square root of a* is a1, of a? is a, of a6 is a3.
The same rule, when the exponent is an odd number,
gives a fractional exponent, thus : the square root of a3 is cfr ;
or, of ab, is a*. So, also, the square root of a, or a1, is a*.
Therefore, we have <?* = Va, equals the square root of a, and
the cube root of a1 = cfr = Va.
427. — Logarithms. — We have seen in the last article the
nature of fractional exponents. Thus the square root of a"
equals eft, which may be put a**. In this way we may have
an exponent of any fraction whatever, as #**. Between the
exponents 2 and 3, we may have any number of fractional
exponents all less than 3 and more than 2. So, also, the
same between 3 and 4, or any other two consecutive num-
bers.
The consideration of fractional exponents or indices has
led to the making of a series of decimal numbers called
logarithms, which are treated in the manner in which expo-
nents are treated ; namely —
To multiply numbers add their logarithms.
To divide numbers, subtract the logarithm of the divisor from
the logarithm of the dividend.
426 ALGEBRA.
To raise any number to a given power, multiply its logarithm
by the exponent of that po^ver.
To obtain the root of any power, divide the logarithm of the
given number by the exponent of the given power.
As an example by which to exemplify the use of loga-
rithms : What is the product of 25 by 375 ?
We first make this statement :
Log. of 25- = i-
" 375* =2.
Putting at the left of the decimal point the integer char-
acteristic, or whole number of the logarithm at one less than
the number of figures in the given number at the left of its
decimal point.
To find the decimal part of the required logarithm we
seek in a book of Logarithms (such as that of Law's, in
Weale's Series, London) in the column of numbers for the
given number 25, or 250 (which is the same as to the man-
tissa) and opposite to this and in the next column we find
7940 and a .place for two other figures, which a few lines
above are seen to be 39 ; annex these and the whole number
is 0-397940. These we place as below :
Log. of 25 • = i • 397940.
Now, to find the logarithm of 375, the other factor, we
turn to 375 in the column ol numbers and find the figures
opposite to it, 4031, which are to be preceded by 57, the two
figures found a few lines above, making the whole, -574031,
which are placed as below, and added together.
Log. of 25- = 1-397940
11 375- = 2-574031
The sum = 3-971971
This sum is the logarithm of the product. To find the
product, we seek in the column of logarithms, headed o-,
for -971971, the decimal part. We find first 97, the first two
EXAMPLES OF LOGARITHMS. 427
figures, and a little below seeking for 1971, the remaining
four figures, we find 1740, those which are the next less,
and opposite these, to the left, we find 7, and above 93, or
together, 937 ; these are the first three figures of the required
product.
For the fourth figure we seek in the horizontal column
opposite 7 and 1740 for 1971, the remaining four figures of
the logarithm, and find them in the column headed 5.
This figure 5 is the fourth of the product and completes
it, as there are only four figures required when the integer
number of the logarithm is 3. The completed statement
therefore is —
Log. of 25-' = 1-397940,
" " 375- = 2-574031,
9375 = 3-97I97I-
Another example in the use of logarithms. What is the
product of 3957 by 94360?
The preliminary statement, as explained in last article, is —
Log- 3957 = 3-
" 94360 = 4-
In the book of logarithms seek in the column of numbers
for 3957. In the first column we find only 395, and opposite
to this, in the next column, we find a blank for two figures,
above which are found 59. Take these two figures as the
first two of the mantissa, or decimal part of the required
logarithm, thus, 0-59. Again, opposite 395 and in the col-
umn headed by 7 (the fourth figure of the given number),
we have the four figures 7366. These are to be annexed to
(o- 59) the first two obtained. The decimal part of the loga-
rithm, therefore, is 0-597366.
To obtain the logarithm for 94360, the other given num-
ber, we proceed in a similar manner, and, opposite 943, we
find 0-97; then, opposite 943 and in column headed 6, we
find 4788, or, together, the logarithm is 0-974788. The
whole is now stated thus —
428 ALGEBRA.
Log. of 3957 = 3-597366
94360= 4-974788
" " 373382000 = 8-572154 = sum of logs.
The two logarithms are here added together, and their sum
is the logarithm of the product of the two given factors.
The number corresponding to the above resultant logarithm
may be found thus: Look in the column headed o for 57,
the first two numbers of the mantissa, then in the same
column, farther down, seek 2154, the other four figures of
the mantissa; or, the four (1709) which are the next less
than the four sought, and opposite these to the left, in the
column of numbers, will be found 373, the first three figures
of the product ; opposite these, to the right, seek the four
figures next less than 2154, the other four figures of the man-
tissa. These are found in the column headed 3 and are
2058. The 3 at the head of the column is the fourth figure
in the product. From 2154, the last four figures of the man-
tissa, deduct the above 2058, or —
2154,
2058,
Remainder, 96.
At the bottom of the page, opposite the next less number
(3727) to that contained in 3733, the answer already found,
seek the number next less to the above remainder, 96. This
is 92-8, and is in the column headed 8. Then 8 is the next
number in the product. From 96 deduct 92-8, and multi-
ply it by 10, or —
96
92-8
3-2 x 10= 32.
Then, in the same horizontal column, seek for 32 or its next
less number. This is 23-2, found in column 2. This 2 is the
next figure in the product. Additional figures may be ob-
tained by the table of proportional parts, but they cannot be
THE SQUARE OF A BINOMIAL. 429
depended upon for accuracy beyond two or three figures.
We therefore arrest the process here.
The product requires one more figure than the integer of
the logarithm indicates; as the integer is 8, there must be
nine figures in the product. We have already six ; to make
the requisite number nine we annex three ciphers, giving
the completed product —
3957 x 9436o = 373382000.
By actual multiplication we find that the true product in the
last article is 373382520. In a book of logarithms, carried to
seven places, the required result is found to be 373382500^
which is more nearly exact.
The utility of logarithms is more apparent when there
are more than two factors to be multiplied, as, in that case,
the operation is performed all in one statement. Thus:
What is the product of 3-75, 432-95, 1712, and 0-0327 ?
The statement is as follows :
Log. 3-75 = 0-574031
432-95 = 2-636438
1712- = 3-233504
•0327 = 8-
Product = 90891. = 4-958521
16
Explanations of working are given more in detail in most of
the books of logarithms.
428. — Completing the Square of a Binomial. — We
have seen in Art. 412 that the square of a binomial (a + &)
equals a" + 2 ab + b* — a trinomial — the first and last terms
of which are each the square of one of the two quantities,
while the second term contains the second quantity multi-
plied by twice the first quantity —
In analytical investigations it frequently occurs that an
expression will be obtained which may be reduced to this
form :
43° ALGEBRA.
a* + mab=f, (nS.)
in which m is the coefficient of the second term, and a and b
are two quantities represented by a and b or any other two
symbols.
A comparison of this expression with the square of a bi-
nomial (112.) contained in Art. 412, shows that the member
at the left comprises two out of the three terms of the square
of a binomial ; as thus —
a* + 2 a b + b*,
but with a coefficient m instead of 2. It is desirable, as will
be seen, to ascertain a proper third term for the given ex-
pression ; or, as it is termed, " to complete the square." The
method by which this is done will now be shown.
A consideration of the above trinomial shows that the
third term is equal to the square of the quotient obtained by
dividing the second term by twice the square root of the
first ; or —
Now a third term to the above binomial, equation (i 18.), may
be obtained by this same rule. For. example —
The rule for the third term then is: Divide the second term
by twice the square root of the first, and square the quotient.
As an example, let it; be required to find the third term
required to complete the square in the expression —
6 n x -f- 4** =/,
in which n and / are known quantities and x unknown.
Putting it in this form—
4_r2 + 6 nx = /,
and dividing by 4, we have —
BINOMIALS CONTINUED. 431
4 4
which reduces to —
Now applying the above rule for finding the third term, we
have —
which is the required third term. To complete the square
we add this third term to both members of the above re-
duced expression, and have —
The member of this expression at the left is the completed
square of a binomial, the two quantities constituting which
are the square roots of the first and third terms respectively ;
or x and f n, and we therefore have —
-2n*+\-H,-
and now taking the square root of both sides of the expres-
sion, we have —
and, by transferring the second quantity to the right mem-
ber, we have —
an expression in which x, the unknown quantity, is made to
stand alone and equal to known quantities.
The process of completing the square is useful, as has
been shown, in developing the value of an unknown quan-
432 ALGEBRA.
tity where it enters into an expression in two forms, one as
the square of the other.
As an example to test the above result, let/ — 256 and
n = 8. Then we have by the last expression for the value
of*—
__
4
= 1/64+36 - 6,
= y ioo — 6,
x = 10 — 6 = 4.
Now this value of x may be tested in the original expres-
sion —
6 nx + 4** = /,
for which we have —
6x8x4 + 4x42 = /,
192 + 64 = /,
the correct value as above.
PROGRESSION.
429. — Arithmetical Progre§§ion. — In a series of num-
bers, as i, 3, 5, 7, 9, etc., proceeding in regular order, in-
creasing by a common difference, the series is called an
arithmetical progression ; the quantity by which one num-
ber is increased beyond the preceding one is termed the
difference. If d represent the difference and a the first term,
then the progression may be stated thus —
Terms— i, 2, 3, 4, 5,
a, a + d, a + 2 d, a + 3 d, a + 4 d, etc.
The coefficient of d is equal to the number of terms preced-
ing the one in which it occupies a place. Thus the fifth
term is a + ^d, in which the coefficient 4 equals the number
of the preceding terms.
From this we learn the rule by which at once to desig-
ARITHMETICAL PROGRESSION. 433
nate any term without finding all the preceding terms. For
the one hundredth term we should have a + 99 d, or, if the
number of terms be represented by n> then the last term
would be represented by—
/ = a + (n — i) d. (H9-)
For example, in a progression where a, the first term, equals
i, d the difference, 2, and n, the number of terms, 90, the last
term will be—
/ = a + (n -- i) d = i + (.90 — i) 2 = 179.
Therefore, to find the last term :
To the first term add the product of the common difference
into the number of terms less one.
By a transposition of the terms in the above expression,
so as to give it this form—
a = / - (n - i)</, (120.)
we have a rule by which to find the first term, which, in
words, is —
Multiply the number of terms less one by the common differ-
ence, and deduct the product from the last term ; the remainder
will be the first term.
By a transposition of the terms of the former expression
to this form —
/ - a = ( n - i) d,
and dividing both members by (n - i), we have —
rf=^-=^. (121.)
n - i '
which is a rule for the common difference, and which, in
words, is —
Subtract the first term from the last, and divide the remain-
der by the number of terms less one ; the quotient will be the com-
mon difference.
Multiplying both members of the equation (121.) by
(n — i) and dividing by d, we obtain—
434 ALGEBRA.
l-a
n-1 = -J-'
Transferring i to the second member, we have —
l-a
n = , + i ; (122.)
which is a rule for finding the number of terms, and which,
in words, is —
Divide the difference between the first and last terms by the
common difference ; to the quotient add unity, and the sum will
be the number of terms.
Thus it has been shown, in equations (119,) (120), (121),
and (122), that when, of the four quantities in arithmetical
progression, any three are given, the fourth may be found.
The sum of the terms of an arithmetical progression may
be ascertained by adding them ; but it may also be had by a
shorter process. If the terms are written in order in a hori-
zontal line, and then repeated in another horizontal line be-
neath the first, but in reversed order, as follows :
i, 3, 5, 7, 9, 11, 13, 15,
15, 13, 11, 9, 7, 5, 3, i,
16, 16, 16, 16, 16, 16, 16, 16,
and the vertical columns added, the sums will be equal. In
this case the sum of each vertical couple is 16, and there are
8 couples ; hence the sum of these 8 couples is 8 x 16 = 128.
And in general the sum will be the product of one of the
couples into the number of couples. It will be observed
that the first couple contains the first and last terms, I and
15 ; therefore the sum of the double series is equal to the
product of the sum of the first and last terms into the
number of terms. Or if 5 be put to represent the sum of
the series, we shall have—
2 S = (a + I) n,
and, dividing both sides by 2—
-S = („ + /)£; (I23.)
GEOMETRICAL PROGRESSION EXPLAINED. 435
Or, in words : The sum of an arithmetical series equals the prod-
uct of the sum of the first and last terms, into half the number
of terms.
430. — Geometrical Progression. — A series of numbers,
such as i, 2, 4, 8, 16, 32, 64, 128, 256, etc., in which any one
of the terms is obtained by multiplying the preceding one
by a constant quantity, is termed a Geometrical Progression.
The constant quantity is termed the common Ratio, and is
equal to any term divided by the preceding one. Thus in
the above example -- or — Or — — 2, equals the common ra-
o 42
tio of the above series. In the series, i, 3, 9, 27, etc., we
have for the ratio—
27 _ 9 _ 3
~9 "3^1 :
which is the common ratio of this series.
A geometrical series may be put thus :
Terms: I, 2, 3, 4;
Progress. : i, i x 3, i x 3 x 3, i x 3 x 3 x 3 ;
or thus —
Terms: I, 2, 3, 4:
Progress. : i, i x 3, i x 3 2, i x 3 8 ;
in which the common ratio, in this case 3, appears in each
term and with an exponent which is equal to the number of
terms preceding that in which it occupies a place.
If the first term be represented by a and the common ra-
tio by r, then the following will represent any geometrical
progression —
a, ar, arz, ar3, #r4, etc. (124.)
For example, let a = 2 and r = 4 ; then the progression
will be —
2, 8, 32, 128, 512, etc.
436 ALGEBRA.
If r = unity, then when a = 2 the progression becomes —
2, 2, 2, 2, 2, etc.
If r be less than unity, then the progression will be a de-
creasing one.
For example, let a •=. 2 and r = -J. Then we have for
the progression —
1 1 L JL
2j Iy 2' 4' 8' 16' e
If the number of terms be represented by n, and the last by
/, then the last term will be —
For example, let n equal 6, then the progression will be —
Terms: i, 2, 3, 4, 5, 6;
Progress.: a, ar, ar*y ar3, ar\ ar*\
in which the .exponent of the last term equals n — i =
6 _ i = 5.
If 5 be put for the sum of a geometrical progression, we
will have —
S = a + ar + arz + ..... + a rn~* + a rll~l.
Multiply each member by ry then—
Sr = ar + ar2 + ..... + a rn~2 + a r*~l + arn.
Subtract the upper line from the lower; then —
S r = ar + ar2 + ..... + arn~2 + a r11'1 + a r*,
S =g + ar+ar2+ ..... + arn~2 + ar*~l _
Sr - s = - a *"* * ~* + ar",
Sr — s = — a + arn,
S (r — i) = — a + a rn = arn — a,
ar" ~~ a
GEOMETRICAL PROGRESSION CONTINUED. 437
The last term (equation (125.)) equals I— arn~l, and since
arn = r x arn~ * = r /, therefore—
Thus, to find the sum of a geometrical progression : Multi-
ply the last term by the ratio ; from tJie product deduct the first
term, and divide the remainder by the ratio less unity.
For example, the sum of the geometrical progression —
•S = i + 3 + 9 + 27 + 81 + 243 +729 = 1093
by actual addition.
To obtain it by the above rule —
rl — a 3 x 729 — I
s==— — : TfST II093'
the correct result.
If there be a decreasing geometrical progression, as i, -J,
•J, ^T, etc., in which the ratio equals ^, the sum will be —
S=i » T + " + ^ + -g7> etc-> to infinity.
Multiply this by 3, and subtract the first from the last —
35=3+1+- + - + — + — + to infinity.
S= i +i + i + ^ + gIf + to infinity.
2 5 = 3» or 5 = if
In a decreasing progression let r, the common ratio, be
represented by — (b less than c), and the first term by a, then
ihe sum will be —
b b2 b3
S=a + a- + a- + a-3+, etc., to infinity.
438 ALGEBRA.
Multiply this by -, and subtract the product from the
above —
^b b d* b*
S- = a- + a- + a-^ + etc., to infinity.
c c c c
b b* l>3
=z a + a — + a—2 + a-j + to infinity.
£ b b* b3
S — — a — \- a — + <2 -j + to infinity.
cere *
Or— s(i --)«*,
For example, let the first term of a geometrical progression
equal 2, and the ratio equal £, then the sum will be—
From this, therefore, we have this rule for the sum of an in-
finite geometrical progression, namely : Divide the first term
by unity less the ratio.
SECTION X.— POLYGONS.
431. — Relation of Sum and Difference of Two Lines. —
Let AB and CD (Fig. 285) be two given lines; make EH
-B
D
E - ] — |-
H J
FIG. 285.
equal to A B, and HG equal to CD-, then E G equals the
sum of the two lines.
Make FG equal to A B, which is equal to EH.
Bisect E G in J ; then, also, J bisects HF\ for—
and —
EH=FG.
Subtract the latter from the former ; then —
EJ-
but—
E
and —
therefore —
Now, E J is half the sum of the two lines, and HJ is half
the difference ; and—
Ey-Hy=EH=AB.
Or : Half the sum of two quantities, minus half their dif-
ference, equals the smaller of the two quantities.
440
POLYGONS.
Let the shorter line be designated by a, and the longer
by b ; then the proposition is expressed by —
_a+b b — a
2 2
(128.)
We also have EJ+JF—EF— CD\ or, half the mm
of two quantities, plus half their difference, equals the larger
quantity.
432. — Perpendicular, in Triangle of Known Side§. —
Let ABC (Fig. 286) be the given triangle, and CE a perpen-
dicular let fall upon A B, the base. Let the several lines of
the figure be represented by the symbols a, b, c, d, g, and f,
as shown. Then, since A EC and BEC are right-angled
triangles, we have (Art. 416) the following two equations,
and, by subtracting one fr.om the other, the third—
Then (Art. 414), by substitution, we have —
(f + f)(f-e) = (« + *)(«-*)•
By division we obtain —
_
a- 6)
f+g
RULE FOR TRIGONS. 441
According to Art. 431, equation (128.), we have —
In this expression let the value of / — g, as above, be
substituted, then we will have—
~=f+* (<* + *) (a - b)
Multiply the first fraction by (f + g), then join the two
fractions, when we will have —
The lines f and g, in the figure, together equal the line
c ; therefore, by substitution —
f - (a + b) (a - b]
g = -^- -i. (129.)
This is the value of the line g.
It may be expressed in words, thus: The shorter of the
two parts into which the base of a triangle is divided by a
perpendicular let fall from the apex upon the base, equals the
quotient arising from a division by twice the base, of the differ-
ence between the square of the base and the product of the sum
and difference of the two inclined lines.
As an example to show the application of this rule, let
a — 9, b = 6, and c = 12 ; then equation (129.) becomes —
12' - (9 + 6) (9 - 6)
2 X 12
. .144 - iT><"3
-
99-
442 POLYGONS.
Now, to obtain the length of d, the perpendicular, by the
figure, we have —
-and, extracting the square root —
or, in words : The altitude of a triangle equals the square root
of the difference of the squares of one of the inclined sides and
its base.
As an example, take the same dimensions as before, then
equation (130.) becomes —
The square of 6 = 36-
" 4i = 17-015625
62-4i2= 18^984375,
the square root of which is 4-44234; therefore
d= t-- 4^ = 4. 44234.
This may be tested by applying the rule to the other in
clined side and its base —
c = 12
* = 4*
/= 71-
Then, ^-
9' = 81-
?%*= 62-015625
9' - 7F = ^8-984375.
TRIGON — RADIUS OF CIRCLES.
443
The same result as before, producing for its square root
the same, 4-44234, the value of d\ therefore—
433. — Trigon : Radiu§ of €ircum§cribed and In§cribed
Circles: Area.— Let A B C (Fig. 287) be a given trigon or
triangle with its circumscribed and inscribed circles. Draw
the lines A D F, DB and D C.
The three triangles, A B D, A CD, and B D C, have their
apexes converging at D, and form there the three angles,
A DB, ADC, and B D C. These three angles together form
four right angles (Art. 335), and each of them, therefore,
equals f of a right angle.
The angles of the triangle BDC together equal two
right angles (Art. 345). As above, the angle BDC equals |
of a right angle, hence 2 — -J = ^^ — | of a right angle,
equals the sum of the two remaining angles at B and C.
The triangle BDC is isoceles (Art. 338); for the two sides
B D and D C, being radii, are equal ; therefore the two angles
at the base B and C are equal, and as their sum, as above,
equals f of a right angle, therefore each angle equals -J of a
right angle. Draw the two lines FC and F B. Now, be-
444 POLYGONS.
cause Z> C and DF are radii, they are equal, hence DFC is
an isoceles triangle.
It was before shown that the angle B D C equals | of a
right angle; now, since the diameter A F bisects the chord
B C, the angles B D E ^nd E D C are equal, and each equals
the half of the angle B D C\ or, \ of f of a right angle equals
£ of a right angle. Deducting this from two right angles
(the sum of the three angles of the triangle), or 2 — f =
\\ — | of a right angle equals the sum of the angles at F and
C', hence each equals the half of f, or f of a right angle;
therefore the triangle DFC is equilateral. The triangles
DBF and DFC are equal. The angles B D C and B F C are
equal; the line BC is perpendicular to D F and bisects it,
making DE and EF equal; hence DE equals half D F, or
DB, radii of the circumscribing circle. Therefore, putting
R to represent B D, the radius of the circumscribing circle,
and b = B C, a side of the triangle A B C, by Art. 416, we
have —
+ DE
Transferring and reducing—
4 " 4'
4/ 4
Ijpt^i^
4 ~4
4 x I^»==i3»==^
34 ~3 ~3 '
Or, The Radius of the circumscribing circle of a regular trigon
or equilateral triangle, equals a side of the triangle divided by
the square root of 3.
AREA OF EQUILATERAL TRIANGLE.
445
By reference to Fig. 287 it will be observed, as was
above shown, that D E = E F= - = - ; or, D E, the ra-
dius of the inscribed circle, equals half the radius of the
circumscribed circle; or, again, dividing equation (131.) by
2, we have —
R _ b
2 " 2 t/y-
and, putting r for the radius of the inscribed circle, we
have —
Or: The radius of the inscribed circle of a regular trigon equals
the half of a side of the trigon divided by the square root of 3.
To obtain the area of a trigon or equilateral triangle ; we
have (Art. 408) the area of a parallelogram by multiplying
its base into its height ; and (Arts. 341 and 342) the area of a
triangle is equal to half that of a parallelogram of equal base
and height, therefore, the area of the triangle BD &(Fig. 287)
is obtained by multiplying B C, the base, into the half of
ED, its height. Or, when A^ is put for the area —
or_ ,:;/;'• *=>*£.,
substituting for R its value (131.)—"
jr=*x-4=
4 1/3
4^3
This is the area of the triangle BD C.
The triangle A B C is compounded of three equal tri-
angles, one of which is the triangle B D C ; therefore the
area of the triangle ABC equals three times the area of the
triangle B DC\ or, when A represents the area —
446
POLYGONS.
4 1/3
(1330
Or: The area of a regular /rz^wz or equilateral triangle
equals three fourths of the square of a side of the triangle di-
vided by the square root of 3.
— Tetragon ; Radius of Circumscribed and In-
scribed Circles: Area. — Let A B CD (Fig. 288) be a given
tetragon or square, with its circumscribed and inscribed
FIG. 288.
circles, of which A E is the radius of the former and EF that
of the latter. The point F bisects A B, the side of the
square. A F equals EF and equals half A B, a side of the
square. Putting R for the radius of the circumscribed
circle and b for A B, we have (Art. 416)—
:7T: <'34-)
Or: The radius of the circumscribed circle of a regular tetra-
gon equals a side of the square divided by the square root of 2.
SIDE AND AREA OF HEXAGON.
447
By referring to the figure it will be seen that the radius
of the inscribed circle equals half a side of the square —
b
r —
(135.)
The. area of the square equals the square of a side —
A = b*. (136.)
435. — Hexagon: Radius of Circumscribed and In-
scribed Circles: Area.— Let A B C D EF(Fig. 289) be an equi-
lateral hexagon with its circumscribed and inscribed circles,
of which EG is the radius of the former, and G H that of
the latter. The three lines, A D, BE, and C F, divide the
FIG. 289.
hexagon into six equal triangles with their apexes converg-
ing at G. The six angles thus formed at G are equal, and
since their sum about the point G amounts to four right
angles (Art. 335), therefore each angle equals £ or f of a
right angle. The sides of the six triangles radiating from G
are the radii of the circle, hence they are equal ; therefore,
each of the triangles is isosceles (Art. 338), having equal angles
at the base. In the triangle EGD, the sum of the three
angles being equal to two right angles (Art. 345), and the
angle at G being, as above shown, equal to f of a right angle,
therefore the sum of the two angles at E and D equals
2 — £ = J of a right angle ; and, since they equal each other,
44^ POLYGONS.
therefore each equals f of a right angle and equals the angle
at G ; therefore E G D is an equilateral triangle. Hence
ED, a side of a hexagon, equals E G, the radius of the circum-
scribing circle —
R=b. (137.)
As to the radius of the inscribed circle, represented by G H,
a perpendicular from the centre upon ED, the base; the
point H bisects E D. Therefore, E H equals half of a side
of the hexagon, equals half the radius of the circumscribing
circle. Let R = this radius, and r the radius of the inscribed
circle, while b = a side of the hexagon ; then we have (Arts.
353 and 416)—
i
r —
Now, R =1 b, therefore —
r=^-- ' (138.)
Or : The radius of the inscribed circle of a regular hexagon
equals the half of a side of the hexagon, multiplied by the
square root of 3.
As to the area of the hexagon, it will be observed that the
six triangles, A B G, B G C, etc., converging at G, the centre,
are together equal to the area of the hexagon. The area of
E G D, one of these triangles, is equal to the product of £ D,
the base, into the half of G H, the perpendicular ; or, when
N is put to equal the area—
G H
SIDE AND AREA OF OCTAGON.
449
and, since rt as above, equals * 3 — ,
2
This is the area of one of the six equal triangles ; therefore,
when A is put to represent the area of the hexagon, we have—
A =
(139.)
Or : The area of a regular hexagon equals three lialves of the
square of a side multiplied by the square root of $.
FIG. 290.
4-36- — Octagon: Radius of Circumscribed and In-
icribed Circles: Area. — Let C E D B F (Fig. 290) represent a
quarter of a regular octagon, in which ^is the centre, ED
a side, and CE and DB each half a side, while CF and
£Fare radii of the inscribed circle, and BF and DF are
radii of the circumscribed circle.
450 POLYGONS.
Let R represent the latter, and r the former ; also let b
represent ED, one of the sides, and n be put for A D, and
for A E. Then we have —
b_
2~ T*
b
or- n = r-->
Since A D Eis a right-angled triangle (Art. 416), we have —
T = ED\
n* = b\
b*
Placing the value of n, equal to the value before found,
we have —
b b
f i i\
-=f — - + - )
2 V 1/2 2/
This coefficient may be reduced by multiplying the first
fraction by ^2, thus—
JL x t5 = .^i
V2 X 2 X>
RULES FOR OCTAGON. 451
therefore —
r =
Or : The radius of the inscribed circle of a regular octagon
equals half a side of the octagon multiplied by the sum of unity
plus the square root of 2. In regard to the radius of the cir-
cumscribed circle, by Art. 416 we have —
In this expression substituting for ra, its value as above, we
have —
The square of the coefficient ( t/2 + i ) by Art. 412 equals
21/2+1 =21/2 + 3, then —
Or : The radius of the circumscribed circle of a regular octagon
equals half a side of the octagon multiplied by the square root of
the sum of twice the square root of 2 plus 4.
In regard to the area of the octagon, the figure shows
that one eighth of it is contained in the triangle D E F.
452 POLYGONS.
The area of D E F, putting it equal to N, is —
B F
N = EDx — -,
N = b x — ,
AT = ( 1/2" + i) — .
4
This is the area of one eighth of the octagon ; the whole
area, therefore, is —
.
4
A = (V~2+i)2b\ (142.)
Or : The area of a regular octagon equals twice the square of a
side, multiplied by the sum of the square root of 2 added to unity.
When a side of the enclosing square, or diameter of the
inscribed circle, is given, a side of the octagon may be found ;
for from equation (140.), multiplying by two, we have —
2 r - ( V~2 + i) b.
Dividing by V.2 + i, gives —
The numerator, 2 r, equals the diameter of the inscribed
circle, or a side of the enclosing square ; therefore :
The side of a regular octagon, equals a side of the enclosing
square divided by the sum of the square root of 2 added to unity.
437.— Dodecagon : Radius of Circumscribed and In-
scribed Circles: Area. — Let A B C (Fig. 291) be an equilat-
SIDE AND AREA OF DODECAGON. 453.
era! triangle. Bisect A B in F\ draw C FD ; with radius A C
describe the arc A D B. Join A and D, also D and B ; bisect
A D in E ; with the radius E C describe the arc E G. Then
A D and D B are sides of a regular dodecagon, or twelve-
sided polygon ; of which A C, D C, and B C are radii of the
circumscribing circle, while E C is a radius of the inscribed
circle.
The line A B is the side of a regular hexagon (Art. 435).
Putting R equal to A C the radius of the circumscribing cir-
cle ; r, = E C, the radius of the inscribed circle ; £, = A D, a
side of the dodecagon, and n — D F. Then comparing the
FIG. 291.
homologous triangles, ADF and A EC (the angle ADF
equals the angle EA C, and the angles DFA and A E C are
right angles); therefore, the two remaining angles DAF
and A CE must be equal, and the two triangles homologous
(Art. 345). Thus we have—
DF : DA : : AE : A C,
n : b : : : R,
**&
454 POLYGONS.
In Art. 435 it was shown that FC (Fig. 291), or G H oi
Fig. 289, the radius of the inscribed hexagon, equals V~l ~,
n
and in which its b = R ; Fc — VJ — .
Now ( Fig. 291) —
= DC - FC,
or —
n = R-^~
Substituting this value of n, in the above expression, we
have —
R-
~
Multiplying by R and reducing, we have
R = |_ ^_ b. (144.)
Or : The radius of the circumscribed circle of a regular dodec-
agon, equals . a side of the dodecagon multiplied by the square
root of a fraction, having unity for its numerator and for its
denominator 2 minus the square root of 3.
Comparing the same triangles, as above, we have —
FD \ FA : : EA ; EC,
or—
R
. b
n : —
2
: : - : r,
Rb
Rb
~ 4 n ~ *
\R(\ -i|/^
b
('450
RULE FOR DODECAGON. 455
Or : The radius of the inscribed circle of a regular dodecagon
equals a side of the dodecagon divided by the difference between
4 and the square root of 3.
The area of a dodecagon is equal to twelve times the area
of the triangle ADC (Fig. 291). The area of this triangle is
equal to half the base by its perpendicular ; or, A E x E C ;
or —
b
or, where N equals the area —
Or, for the area of the whole dodecagon —
12 N — 6 br,
A =6br.
Substituting for r its value as above, we have —
Or : The area of a regular dodecagon equals the square of a
side of the dodecagon, multiplied by a fraction having 6 for its
numerator, and for its denominator, 4 minus twice the square
root of 3.
438. — Hecadecagon : Radius of Circumscribed and Iii-
§cribed Circle§ : Area. — Let A B CD (Fig. 292) be a square
enclosing a quarter of a regular octagon C EFB, E F being
one of its sides, and C E and FB each half a side, while F D
is the radius of the circumscribed circle, and J D the radius
of the inscribed circle of the octagon. Draw the diagonal
A D ; with DFior radius, describe the circumscribed circle
EGF\ join G with F and with E ; then EG and GFvfill
each be a side of a regular hecadecagon, or polygon of six-
teen sides.
An expression for F D, the radius of the circumscribed
456
POLYGONS.
circle, may be obtained thus: Putting FD = R] H D = r;
G F = b\ GJ — n\ and J F — - (Art. 416), we have —
GT = GF* — JF\
•• = »•- (9-
C D
FIG. 292.
Comparing the two homologous (Art. 361) triangles, GJF
and F H D (Art. 374), we have —
Gy : GF : : HF : FD,
n, b :: I : *,
Putting this value of n' in an equation against the former
value, we have—
In Art. 436, the value of F D, as the radius of the cir-
cumscribed circle of a regular octagon, is given in equation
(141.) as—
b
R
V2
SIDE AND AREA OF HECADECAGON. 457
in which b represents a side of the octagon, or E Ft for
which we have put s. Substituting s for b and putting the
numerical coefficient under the radical, equal to B, we
have —
Squaring each member gives—
From which, by transposition, we have —
£2
Substituting in the above expression for (— ) , this value
• \2 /
of it, gives —
^- = t>>-*\
4^2~ B .
Transposing, we have—
-*1 + *'=*-.
4R* B
Multiplying the first term by B, and the second by
we have —
*
Bb* + 4^4 _ ^
Transposing, we have —
2= -Bb\
458 POLYGONS.
To complete the square (Art. 428) we proceed thus —
Taking the square root, we have
Restoring B to its value, 2 I/I + 4 as above, we have—
B — I = 2^2
multiply these —
2 + 2 i/J,
3 +
= 4/2"+ 2.
Therefore—
2. (147.)
RULES FOR HECADECAGON. 459
Or: The radius of the circumscribed circle of a regular
hecadecagon equals a side of the hecadecagon multiplied by the
square root of the sum of two quantities, one of which is the
square root of 2 added to 2, and the other is the square root of
the sum of seven halves of the square root of 2 added to 5.
To obtain the radius of the inscribed circle we have (Fig.
292)—
H D* = FD* — H F\
Substituting for R a its value as above, we have —
r a = &* ( B(B} + * B) - ,
The coefficient of b is the same as in the case above, ex-
cept the — i; therefore its numericaK value will be i less,
or—
r = b \/ ^S + I V2 + V2 + if. (148).
Or: The radius of the inscribed circle of a regular hecadeca-
gon equals a side of the hecadecagon multiplied by the square root
of two quantities, one of which is the square root of 2 added to
if, and the other is the square root of the sum of seven halves of
the square root of 2 added to 5.
To obtain the area of the hecadecagon it will be observed
that the area of the triangle G FD (Fig. 292) equals HD x
H F, and that this is the TV part of the polygon ; we there-
fore have —
A = \6HDxHF,
A = i6r- = Sr&.
2
460 POLYGONS.
The value of r is shown in (148.); therefore we have —
A = 8 b
+ 1 .
+ if.
(H9-)
Or : The area of a regular hecadecagon equals eight times
the square of its side, multiplied by the square root of two qtian-
tities, one of which is the square root of 2 added to if, and the
other is the square root of the sum of seven halves of the square
root of 2 added to 5.
439.— Polygon* : Radius of Circumscribed and I u scribed
Circles : Area. — In Arts. 433 to 438 the relation of the radii
to a side in a trigon, tetragon, hexagon, octagon, dodeca-
gon and hecadecagon have been shown by methods based
upon geometrical proportions. This relation in polygons
of seven, nine, ten, eleven, thirteen, fourteen and fifteen
sides, cannot be so readily shown by geometry, but can be
easily obtained by trigonometry — as also said relation of the
parts in a regular polygon of any number of sides. The na-
ture of trigonometrical tables is discussed in Arts. 473 and
474. So much as is required for the present purpose will
here be stated.
Let ABC (Fig. 293) represent one of the triangles into
which any polygon may be divided, in which B C = b = a
side of the polygon ; A C — R = the radius of the circum-
scribed circle ; and A D = r = the radius of the inscribed
circle.
GENERAL RULES FOR POLYGONS. 461
Make E C equal unity ; on C as a centre describe the arc
E F\ draw FH and E G perpendicular to B C, or parallel to
A D ; then for the uses of trigonometry E G is called the
tangent of c, or of the angle A C B, and FHis the sine, and
H C the cosine of the same angle.
These trigonometrical quantities for angles varying from
zero up to ninety degrees have been computed and are to be
found in trigonometrical tables.
Referring now to Fig. 293 we have —
HC : FC : : DC : AC,
b
cos. c : I : : — : R,
(150.)
Again —
E C : E G : : D C : A D,
b
I : tan. c : : - : r,
r = -• tan. c. (151.)
These two equations give the required radii of the cir-
cumscribed and inscribed circles. They may be stated thus :
The radius of the circumscribed circle of any regular poly-
gon equals a side of the polygon divided by twice the cosine of
the angle formed by a side of the polygon and a radius from one
end of the side.
The radius of the inscribed circle of any regular polygon
equals half of a side of the polygon imiltiplied by the tangent of
the angle formed by a side of the polygon and a radius from one
end of the side.
The area of a polygon equals the area of the triangle
ABC (Fig. 293), (of which B C is one side of the polygon
and A is the centre), multiplied by the number of sides in
the polygon ; or, if n be put to represent the number of the
sides and A the area, then we have—
POLYGONS.
A = Bn,
in which B equals the area of the triangle. The area of A
B C (Fig. 293) is equal to AD x B D, or —
For r substituting its value, as in equation (151.), we have —
b b i 7
B = - tan. c- = — b" tan. c.
2 24
Therefore, by substitution —
A =-b*n tan.*. (152.)
Or : The area of a regular polygon equals the square of a
side of the polygon, multiplied by one fourth of the number of
its sides, and by the tangent of the angle formed by a side of the
polygon, and a radius from one end of the sides.
440. — Polygons : Their Angles. — Let a line be drawn
from each angle of a regular polygon to its centre, then
these lines form with each other angles at the centre, which
taken together amount to four right angles, or to 360 de-
grees (Arts. 327, 335).
If this 360 degrees be divided by the number of the sides
of the polygon, the quotient will equal the angle at the cen-
tre of the polygon, of each triangle formed by a side and two
radii drawn from the ends of the side. For example: if
ABC (Fig. 293) be one of the triangles referred to, having
B C one of the sides of the polygon and the point A the cen-
tre of the polygon, then the angle B A (Twill be equal to 360
degrees divided by the number of the sides of the polygon.
If the polygon has six sides, then the angle B A C will contain
= 60 degrees ; or if there be 10 sides, then the angle at
A, the centre, will contain - ---- = 36 degrees. The angle
SIDE AND AREA OF PENTAGON. 463
BAD equals half the angle B A C, or, when n equals the
number of sides, the angle BAG equals —
360
»
n
B A C
and the triangle B A D = , equals —
360
2 n
Now the angles B A D + D B A equal one right angle
(Art. 346), or 90 degrees. Hence the angle DBA =90° -
BAD,or the angle c equals —
(1530
2 n
For example, if n equal 6, or the polygon have six sides,
then—
Therefore, the angle c, contained in equations (150.), (151.)*
and (152.), equals 90 degrees, less the quotient derived from a di-
vision of 360 by twice the number of sides to the polygon.
441. — Pentagon: Radius of the Circumscribed and In-
scribed Circles: Area.— The rules for polygons developed
in the two former articles will here be exemplified in their
application to the case of a regular pentagon, or polygon of
five sides.
To obtain the angle c° (153.), we have n = 5, and —
,•>= 90°- 3g = go-36 = 54°.
For the radius of the circumscribed circle, we have
(150.)-
2 COS. C
464 POLYGONS.
b
2 cos. 54C
i
2 cos. 54°
Using a table of logarithmic sines and tangents (Art. 427),
we have —
Log. 2 =0-3010300
Cos. 54° = 9-7692187
Their sum = 0-0702487 — subtracted from
Log. i = o-ooooooo
0-85065 =9-9297513
Therefore —
,£ = 0-85065 £.
Or : The radius of the circumscribed circle of a regular/^ta-
gon- equals a side of the pentagon multiplied by the decimal o • 8 5065 .
For the radius of the inscribed circle, we have (151.)—
= - tan. ct
.tan. 54°
r = o
For this we have —
Log. tan. 54° = 0-1387390
Log. 2 = 0-3010300
0-68819 = 9-8377090.
Therefore —
r = 0-68819 b.
Or: The ra&usofthe inscribed circle of a regular pentagon
equals a side of the pentagon multiplied by the decimal 0-68819.
For the area we have (152.)—
A =%fr*n tan. c,
A = J x 5 tan. 54° b\
A =ftan. 54° b\
TABLE FOR REGULAR POLYGONS. 465
For this we have —
Log. 5. = 0-6989700
Log. tan. 54° =_- o- 1387390
0-8377090
Log. 4 = 0-6020600
1-72048 — 0-2356490
Therefore—
A = i- 72048 b \
Or: The area of a regular pentagon equals the square of its
side multiplied by I • 72048.
1-42. — Polygons Table of €on§tant multipliers. — To
obtain expressions for the radii of the circumscribed and in-
scribed circles, and for the area for polygons of 7, 9, 10, n,
13, 14, and 15 sides/a process would be needed such pre-
cisely as that just shown in the last article for a pentagon,
except in the value of n and c, which are the only factors
which require change for each individual case.
No useful purpose, therefore, can be subserved by ex-
hibiting the details of the process required for these several
polygons. The values of the constants required for the
radii and for the areas of these polygons have been com-
puted, and the results, together with those for the polygons
treated in former articles, gathered in the annexed Table of
Regular Polygons.
REGULAR POLYGONS.
SIDES.
R
b .
r
1> =
A
b~*~
o Trie-on
• ^77^
•28868
•4TJOI
A Tetragon
•7O7II
• 5OOOO
I -OOOOO
5 Pentagon
.8co6t;
•68819
I • 72O48
•ooooo
. 86603
2. cr\So8
•152-18
I 03826
3-6T3QI
8 Octagon
* 30656
I • 2O7 I I
4- 8284^
• 4.6 1 QO
I •37'374
6- 18182
10 Decagon ... .
•61803
i -^884
II. Undecagon
• 7747^
i • 70284
<J94^J-
9. a6c6d.
12 Dodecagon
. Q^lS^
i • 86603
13. Tredecagon
2 • 08020
2-02858
11 ' I 8^77
14 Xetradecagon ... .
2 • 24608
2 • I 0064
15. Pentadecagon
2 • 4OJ.8 7
2 • ^^2^1
XD JJ451
17 • 6j.2^6
16 Hecadecagon
2 • 56202
2. e T'l67
466 POLYGONS.
In this table R represents the radius of the circumscribed
circle ; r the radius of the inscribed circle ; b one of the
sides, and A the area of the polygon. By the aid of the
constants of this table, R, the radius of the circumscribed
circle of any of the polygons named, may be found when a
side of the polygon is given. For this purpose, putting m
for any constant of the table, we have —
R = bm. (1S4-)
As an example : let it be required to find R, for a penta-
gon having each side equal to 5 feet ; then the above expres-
sion becomes —
R — 5 x 0-85065,
R = 4-25325-
The radius will be 4 feet 3 inches and a small fraction.
In like manner the radius of the inscribed circle will be —
r = bm; (1550
and for a pentagon with sides of 5 feet, we have —
r = 5 x 0-68819,
r — 3-44095-
Or, the radius of the inscribed circle will be 3 ft. -j^ and a
small fraction. Or, multiplying the decimal by 12, 3 ft. 5 in.
-f^Q and a small fraction.
The area of any polygon of the table may be obtained
by this expression—
A^b*m; (156.)
and, applying this to the pentagon as before, we have —
A — 5 2 x i • 72048,
A — 43-012.
EXPLANATION OF THE TABLE. 467
Or, the area of a pentagon having its sides equal to 5 feet,
is 43 feet and T-^f7 of a foot.
By the constants of the table a side of any of its poly-
gons may be found, when either of the radii, or the area,
are known.
When R is known, we have —
When r is known, we have-
When the area is known, we have —
SECTION XL— THE CIRCLE.
443. — Circles: Diameter and Perpendicular: mean
Proportional. — Let ABC (Fig. 294) be a semicircle. From
C, any point in the curve, draw a line to A and another to
B\ then ABC will be a right-angled triangle (Art. 352).
Draw the line CD perpendicular to the diameter AB\
then C D w\\\ divide the triangle A B C into two triangles,
A CD and C B D, which are homologous. For, let the
triangle C B D be revolved on D as a centre until its line
CD shall come to the position E D, and the line DB oc-
cupy the position D F, each in a position at right angles
to its former position, the point B describing the curve
B F, and the point C the curve C E, and each forming a
quadrant or angle of ninety degrees. Since these points
have revolved ninety degrees, therefore the three lines of
the triangle CBDhave revolved into a position at right
angles to that which they before occupied ; hence the line
EFis at right angles to CB} and (from the fact that A C B is
a right angle) parallel with A C, Since the triangle EFD
equals the triangle C B D. and since the lines of E FD are
parallel respectively to the corresponding lines of A CD,
therefore the triangles^ CD and C B D are homologous.
Comparing the lines of these triangles and putting a =
A B, y = C D, and x — D B, we have —
RADIUS FROM CHORD AND VERSED SINE. 469
DB : D C : : D C : A D,
x : y : : y : a — x,
y* = x (a — x], (160.)
Or, in a semicircle, a perpendicular to the diameter terminated
by the diameter and the curve is a geometric mean, or mean
proportional, between the two parts into which the perpendicular
divides the diameter.
444. — Circle : Radius from Given Chord and Versed
Sine. — Let A B (Fig. 295) be a given chord line and CD a
versed sine. Extend CD to the opposite side of the circle ;
it will pass through F, the centre. Join A and C, also E and
B
B. The line A D, perpendicular to the diameter C E, is
a mean proportional between the two parts CD and DE
(Art. 443) ; or, putting a = A D, b = C D, and r equal the
radius FE, we have —
C D \ AD \\ AD \ DE\
b : a : : a : 2 r — b,
-2 rb-b*,
+ l>* = 2rb,
r -
(161.)
470
THE CIRCLE.
Or : The radius of a circle equals the sum of the squares of half
the chord and the versed sine, divided by twice the 'versed sine.
Another expression for the radius may be obtained ; for
the two triangles C B D and C E B (Fig. 295) are homologous
(Art. 443) and their corresponding lines in proportion. Put-
ting/for CB, we have —
or —
or —
and —
CD : CB : : CB : C E,
v :/::/: 2 r,
f ' — 2 rv,
r =
2V
(162.)
Or : The radius of a circle equals the square of the chord of half
the arc divided by twice the versed sine.
4-45. — Circle: Segment from Ordinate§. — When the curve
of a segment of a circle is required for which the radius can-
not be used, either by reason of its extreme length, or be-
FIG. 296.
cause the centre of the circle is inaccessible, it is desirable
to obtain the curve without the use of the radius. This may
be done by calculating ordinates, a rule for which will now
be developed.
Let DC B (Fig. 296) be a right angle, and A DB a cir-
cular arc described from C as a centre, with the radius
B C= CD = CP. Draw PM parallel with DC, and A G
parallel with C B. Now, in the segment A D G, we have
given A G, its chord, and D E, its versed sine, and it is re-
RULE FOR ORDINATES. 471
quired to find an expression by which its ordinates, as P F,
may be computed. From Art. 416, we have —
PM*=CP*-CJf*<,
or, putting for these lines their usual symbols —
now we have —
EC= FM,
FM=DC—DE,
FM =r-b.
Then we have —
or, putting t for P F and substituting for PM and FM their
values as above, we have —
t = y-(r-b\
and for y, substituting its value as above, we have—
/r* -x* -(r-b). (163.)
Or: The ordinate in the segment equals the square root of the
difference of the squares of the radius and the abscissa minus
the difference of the radius and the versed sine.
For example : let the chord A G (Fig. 296) in a given case
equal 20 feet, and the versed sine, b, or the rise D E, equal 4
feet ; and let the ordinates be located at every 2 feet along
the chord line, A G.
In solving this problem we require 'first to find the radius.
This is obtained by means of equation
2b
4/2 THE CIRCLE.
For a, half the chord, we have 10 feet ; for b, the versed
sine, we have 4 feet ; and, substituting these values, we
have —
The radius equals — H'5
The versed sine equals — 4-0
(r-b}= 10-5
The square of 14-5, the radius, equals 210-25. Now we
have, substituting these values in equation (163.) —
-~- 1210-25 — x^ — 10-5.
The respective values of x, as above required, are o, 2,
4, 6, 8 and 10 Substituting successively for x one of these
values, we shall have, when —
x— o; t -- y 210-25 — o2 — 10-5 = 4.
x- 2 ; /-= |/ 210-25 — 2 2 — 10- 5 = 3-8614
4; '=- V 210-2$ — 4*-- 10-5 = 3-4374
# = 6; / = |/ 210-25 — 62 — 10-5 =: 2-7004
* = 8 ; / = 4/ 210-25 — 82 — 10- 5 = i • 5934
r I0' * -: 1/210-25 — io2-- 10-5 = o-o
Values for / may be taken at points as numerous as desira-
ble for accuracy.
In ordinary cases, however, they need not be nearer than
in this example.
After the points are secured, let a flexible piece of wood
be bent so as to coincide with at least four of the points at a
time, and then draw the curve against the strip.
446. — Circle : Relation of Diameter to Circumference.
—In Art. 439 it is* shown that the area of a polygon equals
the radius of the inscribed circle multiplied by half of a
side of the polygon and by the number of the sides ; or,
TO FIND THE CIRCUMFERENCE. 473
A = r x — n = b n ; or, the area equals half the radius by a
2 2
side into the number of sides ; or, half the radius into the
periphery of the polygon. Now, if a polygon have very
small sides and many of them, its periphery will approxi-
mate the circumference of the circle inscribed within it ; in-
deed when the number of sides becomes infinite, and conse-
quently infinitely small, the periphery and circumference
become equal. Consequently, for the area of the circle, we
have —
A = r--c, (164.)
where c represents the circumference.
By computing the area of a polygon inscribed within a
given circle, and that of one circumscribed about the circle,
the area of one will approximate the area of the other in
proportion as the number of the sides of the polygon are
increased.
For example : if polygons of 4 sides be inscribed within
and circumscribed about a circle, the radius of which is I,
the areas will be respectively 2 and 4. If the polygons have
1 6 sides, the areas are each 3 and a fraction, the fractions
being unlike; when they have 128 sides the areas are each
3 • 14 and with unlike fractions ; when the sides are increased
to 2048, the areas each equal 3-1415 and unlike fractions,
and when the sides reach 32768 in number the areas are
equal each to 3-1415926, having like decimals to seven
places. The computations have been continued to 127
places (Gregory's " Math, for Practical Men "), but for all
possible uses in building operations seven places will be found
to be sufficient. From this result we have the diameter in
proportion to the circumference as i : 3- 1415926, or as —
I : 3
i : 3
1:3- 1416.
Of these proportions, that one may be used which will give
474 THE CIRCLE.
a result most nearly approximating the degree of accuracy
required. For many purposes the last proportion will be
sufficiently near the truth.
For ordinary purposes the proportion 7 : 22 is very use-
ful, and is correct for two places of decimals; it fails in the
third place.
The proportion 113 : 355 is correct to six places of deci-
mals.
For the quantity 3-1415926 putting the Greek letter n
(called py\ and 2 r = d for the diameter, we have—
c — n d. (165.)
To apply this : in a circle of 50 feet diameter, what is
the circumference ?
c = 3-1416 x 50
c = 1 57-08 ft.
If the more accurate value of n be used, we have —
c = 3-1415926 x 50,
c = i 57- 07963.
The difference between the two results is 0-00037, which
for all ordinary purposes, would be inappreciable.
By the rule of 7 : 22, we have —
c = 5ox-3T2- _ 157.1428571,
an excess over the more accurate result above, of 0-0632271,
which is about £ of an inch.
Bv the rule of 113 : 355, we have —
c = 50 x fff = 157-079646.
This result gives an excess of only 0-000016; it is sufficiently
near for any use required in building.
From these results we have these rules, namely : To
obtain the circumference of a circle, multiply its diameter by
TO FIND THE AREA. 475
22, and divide the product by 7 ; or, more accurately, multiply
the diameter ^355 and divide the product by 113; or, by mul-
tiplication only, m ultiply the diameter by 3-1416; or, by
3-14159^; or, by 3-1415926; according to the degree of
accuracy required.
And conversely: To obtain the diameter from the cir-
cumference, multiply the circumference by 7 and divide the
product by 22 ; or, multiply by 113 and divide by 355 ; or, di-
vide the circumference by 3-1416; or, by 3-14159^; or, by
3-1415926.
4-47. — Circle : Length of an Arc. — Considering the cir-
cle divided into 360°, the length of an arc of one degree in
a circle the diameter of which is unity may be thus found.
The circumference for 360° is 3- 14159265 ;
3. 14159265 = 0.oo8726fi4625;.
which equals an arc of one degree in a circle having unity
as its diameter; or, for ordinary use the decimal 0-008727
or 0-0087^ may be taken ; or putting a for the arc and g for
the number of degrees, we have —
a = 0-00872665 dg. (166.)
Wherefore : To obtain the length of an arc of a circle,
multiply the diameter of the circle by the number of degrees in
the arc, and by the decimal 0-0087^, or, instead thereof, by
0-008727.
4.43. — Circle: Area. — The area of a circle may be ob-
tained in a manner similar to that for the area of polygons
(Art. 439), in which A—Bn\ B — r — , or—
A = % b n r,
where b equals a side of the polygon and n the number of
sides ; so that b n equals the perimeter of the polygon.
Now, if for the perimeter of the polygon there be sub-
476 THE CIRCLE.
stituted the circumference of the circle, we shall have, put-
ting for the circumference 3- 1416 dy or, n d (Art. 446)—
A = \n dr,
in which r is the radius. Since 2 r — d, the diameter, and
r = -, we have —
• d
And since —
^ = 3.14159265,
\7t = 0-78539816,
or—
\ n = 0-7854, nearly.
Therefore—
(167.)
Or: The area of a circle equals the square of the diameter mul-
tiplied by 0-7854.
B
As an example, the area of a circle 10 feet in diameter is
found thus —
IOX IO = IOO.
100x0-7854 = 78 -54 feet.
449. — Circle: Area of a Sector. — The area of A B CD
(Fig. 297), a sector of a circle, is proportionate to that of the
whole circle. For, as the circumference of the whole circle
is to its area, so is the arc A B C to the area of A B C D.
AREA OF SECTOR. 477
The circumference of a circle is (165.) C= ir d. The area
of a circle is (167.) A = -7854 d*. For the arc ABC put a,
and for the area of A B CD put s. Then we have from the
above-named proportion—
7t d :
_
•J -
*
Tt d
The coefficient 0-7854 is J- (^4rA 448).
4
Therefore, multiplying the fraction by 4, we have —
5- *d\
O - 7 ft »
4 TTdf
or— S = \da = \ra. (168.)
Wherefore : To obtain the #raz of a sector of a circle,
multiply a quarter of the diameter by the length of the arc.
Thus: let A D equal 10; also let A B C = a, equal 12.
Then the area of A £ CD is—
S =%x lox 12,
S = 6o.
The length of the arc may be had by the rule in Art. 447.
450. — Circle: Area of a Segment. — In the last article,
A BCD (Fig. 297) is called the sector of a circle. Of this
the portion included within A E C B is a segment of a circle.
The area of this equals the area of the sector minus the area
of the triangle A D C ; or, putting M for the area of the seg-
ment, S for the area of the sector, and T for the area of the
triangle, then—
M=S- T.
Putting c for A C (Fig. 297) and h for D E, then T = ~ h.
In the last article, s — £ ra, in which a = the length of the
478
THE CIRCLE.
arc ABC.
have —
Substituting this value of s in the above, we
ar —
Or : When the length of the arc is known, also that of the
chord and the perpendicular from the centre of the circle,
then the area of the segment equals the difference between the
product of half the arc into the radius, aud half the chord into
its perpendicular to the centre of the circle.
But ordinarily the length of the arc and of the chord are
unknown. If in this case the number of degrees contained
between the two radii, DA,DC>wz known, then the area of
the segment may be found by a rule which will now be de-
veloped.
In Fig. 298 (a repetition of Fig. 297) upon D as a centre,
and with D F = unity for a radius, describe the arc H F.
Then GFis the sine of the angle C D B, and D G is the co-
sine ; and we have —
or —
Again —
or —
DF : GF : : DC : EC,
I : sin : : r : - = r sin.
DF : DG : : DC : D Ey
i : cos : : r : // = r cos.
RULE FOR AREA OF SEGMENT. 479
By equation (166.) we have —
a = 0-00872665 dg,
in which a is the length of the arc ; g the number of degrees
contained in the arc ; and d is the diameter of the circle.
Since d = 2 r, therefore —
a = 0-0174533 rg.
Putting B for the decimal coefficient, we have —
a = Br g.
The expression (169.), by substitution of values as above,
becomes —
a c
M = -r h.
2 2
B rg
M = — r — r sin. x r cos.
M — \ B gr* — sin. cos. r2
M = r* (%*Bg — sin. cos.)
M — r * (o • 00872665 g — sin. cos.) ( 1 70.)
Or : The area of a segment of a circle equals the square of the
radius into the difference between 0-00872665 times the number
of degrees contained in the arc of the circle, and the product of
the sine and cosine of half the arc.
When the number of degrees subtended by the arc is
unknown, or tables of sines and cosines are not accessible,
then the area may be obtained by equation (169.), provided
the chord and versed sine are known ; but before this equa-
tion can be used, for this purpose, expressions giving their
values in terms of the chord and versed sine must be ob-
tained, for a, the arc, r, the radius, and h, the perpendicular
to the chord from the centre of the circle.
For the value of the arc we have (from " Penny Cycl.,"
Art. Segment] as a close approximation —
480 THE CIRCLE.
By equation (162.) we have —
= 2^'
Then—
h = r — v,
or —
h = f--v.
2 V •
Substituting these values in equation (169.) we have —
This rule is the rule (169.) expanded.
The written rule for equation (169.) may be used, substi-
tuting for " half the arc" one sixth of the difference between
eight times the chord of half the arc and the chord (or \ of 8
times A £>, Fig. 298, minus A C, the chord). Also substitute
for " the radius" the square of the chord of half the arc divided
by twice the versed sine. Also, tor. "its perpendicular to the
centre of the circle" substitute, the quotient of the square of the
chord of half the arc divided by twice the versed sine, minus
the versed sine.
When the arc is small the curve approximates that of a
parabola. In this case the equation for the area of the par-
abola, which is quite simple, may be used. It is this —
Or, in segments of circles where the versed sine is small in
comparison with the chord, the area equals approximately two
thirds of the chord into the versed sine.
SECTION XII.— THE ELLIPSE.
451.— Ellipse : Definitions.— Let two lines, PF, PF' (Fig.
299), be drawn from any point P to any two fixed points
FF'y and let the point P move in such a manner that the sum
of the two lines, PF, PF', shall remain a constant quantity ;
then the curve P M KO G A D B P, traced by P, will be an
Ellipse ; the two fixed points F, F' , the Foci ; the point C at
FIG. 299.
the middle of FF', the centre ; the line A M drawn through
F F' and terminated by the curve, the Major or Transverse
Axis ; the line B O, drawn through C and at right angles to
A M, the Minor or Conjugate Axis; the line G P, drawn
through Pand C and terminated by the curve, the Diameter
to the point P; the line D K drawn through C, parallel with
the tangent P T, and terminated by the curve, the diameter
Conjugate to P G\ the line EH R drawn parallel with D K
is a double ordinate to the abscissas G H "and H Poi the di-
ameter GP(EH = HR) ; the line JL drawn through Fat a
482
THE ELLIPSE.
right angle to A M and terminated by the curve, the Param-
eter, or Latus Rectum.
When the point P reaches and coincides with B, the two
lines PFand PF' become equal.
The proportion between the major and minor axes de-
pends upon the relative position of F,Ff, the foci ; the nearer
these are placed to the extremities of the major axis the
smaller will the minor axis be in comparison with the major
axis. The nearer F, F' approach C, the centre, the nearer
will the minor axis approach the length of the major axis.
When F, F' reach and coincide with the centre, the minor
axis will equal the major axis, and the ellipse will become a
circle. Then we have PF = PF' = B C= A C. From this
we \earnPF+PF'=2A C=AM-t also, when PF= PF',
thenPF=£F=AC.
From this we may, with given major and minor axes,
find the position of F and Ff . To do this, on B, as a centre,
with A C for radius, mark the major axis at F and F' .
452. — Ellip§e : Equations to the Curve. — An equation to
a curve is an expression containing factors two of which,
called co-ordinates, measure the distance to any point in the
curve. For example : in a circle it has been shown (Art.
443) that P N is a mean proportional to A A^and N B. Or,
putting x — A N, y = PN, and a — A B, we have —
AN : PN : : PN : NBy
or —
or —
x : y : : y : a — x,
y a — X (a — X}.
EQUATION TO THE ELLIPSE. 483
This is the equation to the circle having the origin of x
and y, the co-ordinates at A, the vertex of the curve. It will
be observed that the factors are of such nature in this equa-
tion, that it may be employed to measure the distance, rect-
angularly, to (*, wherever in the curve the point P may be
located. By this equation the rectangular distance to any
and every point in the curve may be measured ; or, having
the curve and one of the lines ;ror y, the other may be com-
puted.
From this example, the nature and utility of an equation
to any curve may be understood. The equation to the
ellipse having the origin of co-ordinates at the vertex, is
similar to that for the circle. In the form usually given by
writers on Conic Sections, it is —
in which a — 'A C (Fig. 299) ; b = B C\ x equals A N, and y =
PN.
If, as before suggested, the loci be drawn towards the cen-
tre and finally made to coincide with it, the minor axis would
then become equal to the major axis, changing the ellipse into
a circle. In this case, the factors a and b in the equation would
become equal; and the fraction —5- would equal — , = i,and
a a
hence the equation would become —
or—
y a = x (2 a — x) ;
precisely the same as in the equation to the circle above
shown. The 2 a of this equation is equivalent to a of the
circle ; for a in the ellipse represents only half the major
axis ; while in the equation to the circle a represents the
diameter. The relation between the ellipse and the circle
is thus shown ; indeed, the circle has been said to be an
ellipse in its extreme conditions.
484
THE ELLIPSE.
453.— Ellipse : Relation of Axi§ to Abscissas of Axe§
Multiplying equation (173.) by a* we have—
a* y* = b* (2 ax-x*\
or— #2j/2 = b*x(2 a — x). ^
These four factors may be put in a proportion, thus —
rfa : b* : : x (2 a — x) : y\
representing —
A~C*
7TC
NX NM : P N\
Or : The rectangle of the two parts into which the ordinate
divides the axis major is in proportion to the square of the
ordtnate, as the square of the semi-axis major is to the square
of the semi-axis minor.
It is shown by writers on Conic Sections that this rela-
tion is found to subsist, not only with the axes and ordinate,
but also between an ordinate to any diameter and the ab-
scissas of that diameter ; for example, referring to Fig. 299 —
If A B' P'M (Fig. 301) be a semi-circle, then (Art. 443)
* = A
Substituting this value of A NX N Min —
TC* : ~B~C* - v^Wa : ~PN*>
we have —
A C i BC : : P'N : PN\
RELATION OF TANGENT TO AXIS. 485
Or : The ordinate in the circle is in proportion to its correspond-
ing ordinate in the ellipse, as the semi-axis major is to the semi-
axis minor, or as the axis major is to the axis minor.
454. — Ellipse : Relation of Parameter and Axe§. — The
equation to the ellipse when the origin of the co-ordinates
is at the centre is, as shown by writers on Conic Sections,
thus —
a* y* = a*b*-b* x'\ (174.)
or — a* y"1 — b* (a* — x' 2).
If x' equal C-F (Fig. 299) then the ordinate will be located
^- '^^
01 THE ^f
.'UNIVERSITY
Then-
This is shown also by the figure.
Substituting in the above this value of a* — x'*, we have —
a*y* = &*&* = b\
From which, taking the square root—
ay = b\
or — a : b : : b : y.
Now y, located at FJ, is the semi-parameter; hence we
have the semi-minor axis a third proportional to the semi-
major axis and the semi-parameter. Or : The parameter is a
third proportional to the two axes of an ellipse.
455. — Cllip§e: Relation of Tangent to the Axes. — Let
T T' (Fig. 301) be a tangent to P, a point in the ellipse ; then,
as has been shown by writers on Conic Sections —
or—
CM : CT :: CN : CM.
486 THE ELLIPSE.
Or : The semi-major axis is a mean proportional between the ab-
scissa C N and C T, the part of the axis intercepted between tJie
centre and the tangent.
This relation is found also to subsist between the similar
parts of the minor axis ; for —
This relation affords an easy rule for finding the point T,
or T' ; for from the above we have —
- CN'
or, putting / for C T, we have —
:?y;y / = £ «75.)
or —
t' = --. (176.)
y
Since the value of t is not dependent upon y nor upon b,
therefore / is constant for all ellipses which may be de-
scribed upon the same major axis A M\ and since the circle
is an ellipse (Art. 452) with equal major and minor axes,
therefore rule (175.) is applicable also to a circle, as shown
in Fig. 301.
The equation (175.) gives the value of / = C T. From
this deducting CN = x' , we have N T, the sub tangent, or—
CT- CN = NT,
t - X> = S ;
or, substituting for t its value in (175.), we have —
Or: The subtangent to an ellipse equals the difference between
the quotient of the square of the semi-major axis divided by tlie
abscissa, and the abscissa ; the origin of the co-ordinates being
at C, the centre.
AXES TO CONJUGATE DIAMETER.
487
456. — Ellipse : Relation of Tangent witli the Foci. — Let
the two lines from the foci to P (Fig. 302), any point in the
ellipse, be extended beyond P. With the radius P F' de-
FIG. 302.
scribe from P the arc F' G, and bisect it in H. Then the
line P T, drawn through H, will be a tangent to the ellipse
Sit P.
This has been shown by writers on Conic Sections. The
construction here shown affords a ready method of drawing
a tangent. And from the principle here given we learn
that a tangent makes equal angles with the lines from the
tangential point to the two foci.
For, because GH= HF', we have the angle F' PH =
HPG. The angles H PG and KPF are opposite, and
hence (Art. 344) are equal ; and since the two triangles
F'PffandKPFare each equal to HPG, therefore F' PH
and KPF are equal to each other. Or: A tangent to an
ellipse makes equal angles with the tivo lines drawn from the
point of tangency to the two foci.
Experience shows that light shining from one focus is
reflected from the ellipse into the other focus. It is for this
reason that the two points F and F' are called foci, the plu-
ral oifoczts, a fireplace.
457.— Ellipse : Relation of Axes to Conjugate Diame-
ter—Parallel with K T (Fig. 302) let D E be drawn through
488 THE ELLIPSE.
C, the centre, and L Q through y, one end of the diameter
from the point P. Parallel with this diameter PJ draw L K
and QR through the extremities of the diameter D E. Then
D E is a diameter conjugate to the diameter PJ, and K R,
R Q, QL, and L K are tangents at the extremities of these
conjugate diameters.
Now it is shown by writers on Conic Sections (Fig. 302)
that—
PC*,
or —
Or : The sum of the squares of the two axes equals the sum
of the squares of any two conjugate diameters.
From this it is also shown that the area of the parallelo-
gram K C equals the rectangle A C x B C', or, that a paral-
lelogram formed by tangents at the extremities of any two
conjugate diameters is equal to the rectangle of the axes.
458. — Ellipse ; Area.— Let E equal the area of an ellipse ;
A the area of a circle, of which the radius a equals the semi-
major axis of the ellipse, and let b equal the semi-minor axis.
Then it has been shown that —
E : A : : b : a,
E=Ab-.
a
The area of a circle (Art. 448) is —
A — \ n dr = TT r*,
and when the radius equals a —
A = n a 2,
This value of A, substituted in the above equation, gives—
E = TTtf2-,
a
E = n ab. (178.)
PRACTICAL SUGGESTIONS.
489
Or: The area of an ellipse equals 3- 141 59^ times the product
of the semi-axes ; or 0-7854 times the product of the axes.
459. — Ellipse : Practical Suggestion*. — In order to de-
scribe the curve of an ellipse, it is essential to have the two
axes ; or, the major axis and the parameter ; or, the major
axis and the focal distance.
If the two axes are given, then with the semi-major axis
for radius, from B (Fig. 299) as centre an arc may be made
at F and F't the foci ; and then the curve may be described
by any of the various methods given at Arts. 548 to 552.
If the major axis only and the parameter are given, then
(Art. 454) since —
*» = ay,
we have —
= Vay.
(I79-)
Or : The semi-minor axis of an ellipse equals the square root
of the product of the semi-major axis into the semi-parameter.
Then, having both of the axes, proceed as before.
If the major axis and the focal distance are given, or the
location of the foci ; then with the semi-major axis for ra-
N N
FIG. 303.
dius and from the focal points as centres, describe arcs cut-
ting each other at B and O (Fig. 299). The intersection of
the arcs gives the limit to B O, the minor axis. With the
two axes proceed as before. Points in the curve may be
found by computing the length of the ordinates, and then
the curve drawn by the side of a flexible rod bent to coin-
cide with the several points..
For example, let it be required to find points in the
curve of an ellipse, the axes of which are 12 and 20 feet ; or
490 THE ELLIPSE.
the semi-axes 6 and 10 feet, or 6 x 12 = 72 inches, and 10 x
12 = 1 20 inches.
Fix the positions of the points N N', etc., along the semi-
major axis C ' M (Fig. 303) at any distances apart desirable.
It is better to so place them that the ordinates when drawn
shall divide the curve B PM'mto parts approximately equal.
If CM be divided into eight parts as shown, these parts
measured from C will be well graded if made equal severally
to the following decimals multiplied by CM. In this case
CM= 120; therefore —
C N — 1 20 x 0-3 = 36- = x'
C N' — 120 x 0-475 = 57- = x'
C Nff = 120 x 0-625 = 75 - = x'
Etc., = 120x0-75 = 90- — xf
120 X 0-85 = 102- = Xf
120 x 0-925 = in - = x'
120x0-975 = 117- =• _x'
1 2O X I-O = 1 2O- — X' .
The equation of the ellipse having the origin of co-ordi-
nates at the centre (Art. 454) is —
or, dividing by a*-
a
or —
or- y=-\/ a*~xr*; (i go.)
in which a and b represent the semi-axes. Substituting for
these their values in this case, we have —
LENGTH OF ORDINATES. 491
Now, substituting in this equation the several values of
x* successively, the values of the corresponding ordinates
will be obtained. For example, taking 36, the first value of
x', as above, we have —
y = 0-6 ^74400 — 36*
y = 68-684;
y =. 0-6 V 14400 — 57 2
y = 63-359;
and so in like manner compute the others.
The ordinates for this case are as follows, viz. :
When x' — o, y — 72-0
" x' — 36, y — 68-684
" £n= 57,^ = 63.359
" x' = 75, y = 56-205
" * = 90, y = 47-624
" X' = 102, J/ = 37-928
" x' = ill, 7 = 27-358
" /=; 117, y = 15-999
" *' = 120, J/ — 0-0.
The computation of these ordinates is accomplished easv
ly by the help of a table of square roots and of logarithms.
For example, the work for one ordinate is all comprised
within the following, viz. :
y — 0-6 1/14400 — 362 = 68-684.
I2O2 = I44OO
36 2 = I2Q6
I3I04 = 4-JI74Q39
Half = 2-0587020
0-6 = 9-7781513
68-684 — 1-8368533.
The logarithm of 13104 = 4-1 174039. The half of this is
the logarithm of the square root of 13104. To the half log-
arithm add the logarithm of c-6; the sum is the logarithm
of 68-684 found in the table (see Art. 427).
SECTION XIII.— THE PARABOLA.
460. — Parabola : Definitions. — The parabola is one of
the most interesting of the curves derived from the sections
of a cone. The several curves thus produced are as fol-
lows : When cut parallel with its base the outline is a circle ;
when the plane passes obliquely through the cone, it is an
ellipse ; when the plane is parallel with the axis, but not in
the axis, it is a hyperbola ; while that which is produced by
FIG. 304.
a plane cutting it parallel with one side of the cone is a
parabola.
Let the lines L M and L N (Fig. 304) be at right angles ;
draw CFB parallel with L M\ make LQ — LF\ draw QB
parallel with LF\ then FB = B Q. Now let the line A L
move from F L, but remain parallel with it, and as it moves
let it gradually increase in length in such manner that the
point A shall constantly be equally distant from the line LM
and from the point F. Then A BP, the curve described by
the point A, will be a semi-parabola. For example, the
lines FB and B Q are equal ; the lines PPand PJ/are equal,
and so of lines similarly drawn from any point in the curve
A B P. Let PNbe drawn parallel with LM ; then for the
EQUATION TO THE CURVE. 493
point P, A Nis the abscissa and N P its ordinate (see Art.
452).
The double ordinate C B drawn through F, the focus, is
the parameter. A F is the focal distance. A is the vertex of
the curve. The line L M is the directrix.
4-6 1. — Parabola : Equation to the Curve. — In Fig. 304
FPN is a right-angled triangle, therefore —
= FP* -
but- FP = .MP=LN=AN+AL\
and— FN= A N— A F.
Therefore —
NP* = A N+AL*- A N- A F* ;
or- ' = *
/ being put for the distance LF= FB (see Art. 452). As
in Arts. 412 and 413, we have —
y* = 2px (181.)
by subtraction. This is the usual equation to the parabola,
in which we have the rule : The square of the ordinate equals
the rectangle of the corresponding abscissa with the param-
eter.
From (181.) we have —
x : y : : y : 2p,
or: 1\\e parameter is a third proportional \.o the abscissa and its
corresponding ordinate.
462. — Parabola : Tangent. — From M, any point in the
directrix, draw a line to Ft the focus (Fig. 305) ; bisect M F
in R, and through R draw U T perpendicular to MF, then
the line T U will be a tangent to the curve. For, draw M D
494
THE PARABOLA.
perpendicular to L V, and from P, the point of its intersection
with the line TU, draw a line to F, the focus ; then, because
fiPis a perpendicular from the middle of MF, MPFis an
isosceles triangle, and therefore the lines MPand FP are
equal, or the point P is equidistant from the focus and from
the directrix, and therefore is a point in the curve.
To show that the line TU touches the curve but does not
pass through it, take £7, any point in the line T U, other than
FIG. 305.
the point P; join [7 to Mand to F. Then, since U is a point
in the line T U, M U F, for reasons above given, is an isosce-
les triangle ; from [/draw U F perpendicular to L V. Now,
if the point £/be also in the curve, the lines Wand U F,
by the law of the curve, must be equal ; but U F, as before
shown, is equal to U M, a line evidently longer than UV\
therefore, it is evident that the point U is riot in the curve.
A similar absurd result will be reached if any other point
than the point U in the line U T be assigned, excepting the
RULE FOR THE TANGENT. 495
point P. Therefore the line T P touches the curve in only
one point, P ; hence it is a tangent.
Parallel with L V, from A , draw A S, the vertical tangent.
Now A S bisects M F or intersects it in the point R. For
the two right-angled triangles FL J/and FA R are homolo-
gous ; and because FA = A Z, by construction, therefore
FR = RM.
Or : The vertical tangent bisects all lines which can be drawn
from the focus to the directrix.
The lines PFand FT are equal ; for the lines MPand
N T being parallel, therefore the alternate angles MPT
and N TPare equal (Art. 345) ; and because the line P T bi-
sects M F, the base of an isosceles triangle, therefore the
angles M P T and FPTafe equal. We thus have the two
angles N TP and FP T each equal to the angle MPT-,
therefore the two angles N TP and FP T are equal to each
other ; hence the triangle P F T is an isosceles triangle, hav-
ing the points T and P equidistant from F, the focus.
Also because the line MFis perpendicular to P T, there-
fore the line M F bisects the tangent PT in the point R.
And because TR = R P, therefore, comparing triangles
TRFund TPO, TF= F O.
The opposite angles MPT and U PD made by the two in-
tersecting lines U T and M D (Art. 344) are equal, and since
the angles M P T and F P T are equal, as before shown,
therefore the angles FP T and U PD are equal.
It is because these two angles are equal, that, in reflectors,
rays of light and heat proceeding from /% the focus, are re-
flected from the parabolic surface in lines parallel with the
axis.
For an equation expressing the value of the tangent, we
have —
Or : The tangent to a parabola equals the square root of tJie
sum of four times the square of the abscissa added to the square
of the ordinate.
496 THE PARABOLA.
463. — Parabola: Subtangeiit. — The line T N (Fig. 305),
the portion of the axis intercepted between T, the point of
intersection of the tangent, and N, the foot of a perpendicu-
lar to the axis from P, the point of contact, is the subtangent.
The subtangent is bisected by the vertex, or TA = A N.
For, the two triangles TRA and TPNare homologous;
and, as shown in the last article, the line M F bisects PTin
R-, or TR = R P.
Therefore, we have —
TR : TA : : TP
TR x TN = TA x TP,
but— TR=±TP;
therefore- J- TP x TN= TA x TP,
| TN = TA.
Or : The subtangent of z. parabola is bisected by the vertex ; or
is equal to twice the abscissa.
And because of the similarity of the two triangles TRA
and TP N, as above shown, we have —
NP= 2AR,
y = 2 A R.
Or : The ordinate equals twice the vertical tangent.
464. — Parabola: Normal and Subnormal. — The line
PO (Fig. 305) perpendicular to P T, is the normal and NO,
the part of the axis intercepted between the normal and the
ordinate, is the subnormal. For the normal, from similar
triangles, we have —
TN : NP : :. TP : P O,
TP
2X
DIAMETER AND RECTANGLE OF BASE. 497
Or : The normal equals the rectangle of the ordinate and tan-
gent, divided by twice the abscissa.
The subnormal equals half the parameter. For (181.) —
or— NP* = 2 FB • A N.
Dividing by 2 A N gives—
I FB = ttf (A'}
In the similar triangles (Art. 443) OPNand PT N, we
have —
NO : NP : : NP : NT,
NO = N~P\
NT
As shown in the previous article, N T — 2 A N; therefore —
(B.)
2AN
Comparing equations (A.) and (B.), we have —
NO = FB.
Or : The subnormal of a parabola equals half the parameter,
a constant quantity for the subnormal to all points of the
curve.
465. — Parabola: Diameters. — In the parabola BAC
(Fig. 306), P D, a diameter (a line parallel with the axis) to
the point P, is in proportion to B D x D C, the rectangle of
the two parts into which the base of the parabola is divided
by the diameter.
This may be shown in the following manner :
DP=EN=EA-NA. (A.)
498
THE PARABOLA.
For EA we have, taking the co-ordinates, for the point
C, (i8i.)~
or-
= X
or—
EC
= EA.
0 C
For N A we have, taking the co-ordinates to the point
/>, (181.)-
or —
= X,
NP'-
or-
(C.)
Using these values (B.) and (C.j in (A.), we have—
=£A -NA,
' NP* EC*-NP*
If / be put for B C and n for D C, then—
TO FIND THE CONSTITUENT PARTS. 499
and-
then (Art. 413)—
ft
or (Art. 415)—
2P
= n(l- n)
2P
Dp_DCxBD
~
Now, since 2 /, the parameter, is constant, we have D P,
the diameter, in proportion to D C x B D, the two parts of
the base.
Putting d for the diameter, we have —
(183.)
2p
Or : The diameter of a parabola equals the quotient obtained
by dividing \\\Q rectangle — formed by the two parts into which
the diameter divides the base— by the parameter.
It has been shown by writers on Conic Sections that a
diameter, P J (Fig. 307), to any point Pin a parabola bisects
all chord lines, SG,D£, etc., drawn parallel with the tan-
gent to the point/*; the diameter being parallel with the axis
of the parabola.
466. — Parabola : Elements. — From any given parabola,
to find the axis, tangent, directrix, parameter and focus,
draw any two parallel lines or chords, SG and D E (Fig.
307), and bisect them in H and J\ through these points
draw JP\ then JP will be a diameter of the parabola — a
5oo
THE PARABOLA.
line parallel with the axis. Perpendicular to P J draw the
double ordinate PQ and bisect it in N\ through N and par-
allel with PJ draw TO, cutting the curve in A\ then TO
will be the axis. Make AT- AN, join T and P\ then TP
will be the tangent to the point P\ from P draw PO per-
pendicular to P T\ then P O will be the normal, and NO the
subnormal.
With NO for radius, from N as a centre, describe the
quadrant O R ; draw R C parallel with A O, cutting the curve
FIG. 307.
in C\ from C draw C B perpendicular to A O, cutting A O in
F; then ^will be the focus and C B the parameter. Make
A L — A F; draw L M perpendicular to TO\ then LM will
be the directrix. Extend PJ to meet LM at M\ join P and ,
F\ then, if the work has been properly performed, F P will
equal M P.
4-67. — Parabola : Described mechanically. — With N P
(Fig. 308) a given base, and N A a given height, set perpen-
THE CURVE DESCRIBED MECHANICALLY.
501
dicularly to the base, extend N A beyond A, and make A T
equal to NA ; join T and/*; from P perpendicularly to TP
drawP(9; bisect ON in R; make AL and A F each equal
to N R ; through L, perpendicular to L O, draw D E, the di-
rectrix.
Let the ruler CDESbe laid to the line DE, then with
J G H, a set-square, the curve may be described in the fol-
lowing manner :
Placing the square against the ruler and with its edge
FIG. 308.
J H coincident with the line MPy fasten to it a fine cord on
the edge PE, and extend it from P to Fy the focus, and se-
cure it to a pin fixed in F. The cord FP will equal the
edge M P. To describe the curve set the triangle J G H at
M P E, slide it gently along the ruler towards D, keeping the
edge J G in contact with the ruler, and, as the square is
moved, keep the cord stretched tight, holding for this pur-
pose a pencil, as at K, against the cord. Thus held, as the
square is moved the pencil will describe the curve. That
this operation will produce the true curve we have but to
502
THE PARABOLA.
consider that at all points the line FK will equal KJt which
is the law of the curve (Art. 460).
468. — Parabola : De§cribed from Point§. — With given
base, N P (Fig. 309), and given height, A N, to find the points
D, F, M, etc., and describe the curve. Make A T equal to
A N (Art. 462); join T and P', perpendicular to TP draw
FIG. 309.
PO ; make A B equal to twice NO ; take G, any point in the
axis A O, and bisect B G in J ; on y as a centre describe the
semi-circle B C G cutting A L, a perpendicular to BO in C ;
on A C and A G complete the rectangle A C DG. Then D is
a point in the curve. Take H, another point in the axis ;
bisect B H in. K\ on A" as a centre describe the semi-circle
B E H cutting A L in E ; this by E F and H F, gives F, an-
THE CURVE DESCRIBED FROM ARCS. 503
other point in the curve ; in like manner procure M, and as
many other points as may be desired. This simple and
accurate method of obtaining points in the curve depends
upon two well-established equations ; one, the equation to
the parabola, and the other, the equation to the circle. The
line G Dy an ordinate in the parabola, is equal to A C, an
ordinate in the circle B CG; A G, the abscissa of the para-
bola, is also the abscissa of the circle ; in which we have
(Art. 443)-
AG : AC :: AC : A B,
x : y : : y : a — x,
For the parabola, we have (181.) —
y*=2px.
Comparing" these two equations, we have—
x(a — x] = 2px,
a — x = 2p,
or —
EG — A G= 2 p.
By construction A B equals 2 NO, or twice the subnor-
mal ; the subnormal (Art. 464) equals half the parameter.
Hence, twice the subnormal equals the parameter — equals
2 p. Therefore, the method shown in Fig. 309 is correct.
469. — Parabola: Described from Arcg. — Let N P (Fig.
310) be the given base and A TV the given height of the par-
abola. Make A T(Art. 462) equal A N. Join TtoP; draw
PO perpendicular to PT\ bisect N O in R ; make A L and
A F each equal to N R ; then L M, drawn perpendicular to
TO, will be the directrix. Parallel to L M draw the lines
B D, C E, etc., at discretion. Then with the distance B L for
504
THE PARABOLA.
radius, and on F as a centre, mark the line B D with an arc ;
the intersection of the arc and the line will be a point in the
curve (Art. 460). Again, with C L for radius and on .Fas a
centre, mark the line CE with an arc ; this gives another
point in the curve. In like manner, mark each horizontal
FIG. 310.
line from F as a centre by a radius equal to the perpendicu-
lar distance between that line and L M, the directrix. Then
a curve traced through the points of intersection thus ob-
tained will be the required parabola.
4-70. — Parabola : Dc§cribed from Ordinates. — With a
given base, NP(Fig. 311), and height, A N, a parabola may
be drawn through points J, H, G, etc., which are the extrem-
ities of the ordinates B J, C H, D G, etc. ; the lengths of the
ordinates being computed from the equation to the curve,
(181.)-
For any given parabola, in base and height, the value of
THE CURVE FORMED BY ORDINATES.
505
/ may be had by dividing both members of the equation by
2 4;; by which we have —
y
NP*
2AN'
(A.)
from which, NPand A N being known,/ may be computed.
With the value of/, a constant quantity, determined, the
equation is rendered practicable. For, taking the square
root of each member of equation (181.), we have —
y = V 2p x. (B-)
which by computation will produce the value of y, for every
assigned value of x, as A JS, A C, A D, etc.
FIG. 311.
As an example : let it be required to compute the ordi-
nates in a parabola in which the base, N ' P, equals 8 feet, and
the height, A N, equals 10 feet. With these values equation
(A.) as above becomes —
NP*
2AN
8" _64
2 X 10 ~ 2O " ^' *
506 THE PARABOLA.
Then, with this value in (B.) as above, we have, for each or-
dinate —
y ' z '• V 6.4*.
In order to assign values to x, let A N be divided into
any number of parts at B, C, D, etc., say, for convenience in
this example, in ten equal parts ; then each part will equal
one foot, and we shall have the consecutive values of x = i ,
2, 3, 4, etc., to 10, and the corresponding values of y will be
as follows. When —
x = I, y= 4/6. 4 x i =
3=
x = 4, y = V6-4X "4 = 1/ 25-6 = 5 -0596 =
~32~ = 5 -6569 = (etc.),
38-4 = 6- J968 =
"^8 = 6>6933 =
= 7-5^95 =
With these values of y, respectively, set on the correspond-
ing horizontal lines By, C H, DG, E S, etc., points in the
curve y, //, G, S, etc., are obtained, through which the curve
may be drawn. The decimals above shown are the decimals
of a foot ; they may be changed to inches and decimals of an
inch by multiplying each by 12. For example: 12x0-5297
= 6-3564 equals 6 inches and the decimal 0-3564 of an inch,
which equals nearly | of an inch.
Near the top of the curve, owing to its rapid change in
direction and to the approximation of the direction of the
curve to a parallel with the direction of the ordinates, it is
THE CURVE FOUND BY DIAMETERS.
SO/
desirable to obtain points in the curve more frequent than
those obtained by dividing the axis into equal parts.
Instead, therefore, of dividing the axis into equal parts,
it is better to divide it into parts made gradually smaller
toward the apex of the curve— or, to obtain points for this
part of the curve as shown in the following article.
471. — Parabola: Described from Diameters. — Let EC
(Fig. 312) be the given base and A E the given height, placed
perpendicularly to E C. Divide E C in several parts at
pleasure, and from the points of division erect perpendicu-
lars to E C. The problem is to compute the length of these
diameters, as DP, and thereby obtain points in the curve, as
at P. For this purpose we have equation (183.), which gives
B
\
FIG. 312.
the length of the diameters, and in which n equals D C (Fig.
312), / equals twice E C, and / equals half the parameter of
the curve. The value of p is given in equation (A.), (Art,
470), in which y equals EC (Fig. 312), and ^equals A E.
Substituting these symbols in equation (A.), we have—
y> EC' _^
p~~'' 2x ~ 2xAE~ 2 /T'
where b — EC, the base, and h—AE, the height. For
substituting this, its value, in equation (183.), we have—
hn.(2b — n)
(184.)
508 THE PARABOLA.
As an example : let it be required in a parabola in which
the base equals 12 feet and the height 8 feet, to compute the
length of several diameters, and through their extremities
describe the curve. Then h will equal 8, and b 12.
If the base be divided into 6 equal parts, as in Fig. 312,
each part will equal 2 feet. Then we have —
h 8 8 i
b*~ I2*~~ i44~ 18 '
and —
_ h . N
d— -i n (2b-n),
In this equation, substituting the consecutive values of #,
we have, when—
0x24
n= O, d-- -jg- =0
2 X 22
ti — z,
18
4X 20
•M A
/y _...- - . A . A A A
, , 6x18
*-; 6, ^-^g-
8x16
n= 8, ^=-73— =7-ni
12 X 12
n= 12, d= — -- = 8-0
The several diameters, as P ' D, in Fig. 312, may now be
made equal respectively to these computed values of d, and
the curve traced through their extremities.
AREA EQUALS TWO THIRDS OF RECTANGLE. 509
472. — Parabola: Area. — From (181.), the equation to
the parabola, and by the aid of the calculus, it has been
shown that the area of a parabola is equal to two thirds of
the circumscribing rectangle. For example : if the height,
A E (Fig. 312), equals 8 feet, and EC, the base, equals 12
feet, then the area of the part included within the figure
APC E A equals f of 8x 12 = 1x96 = 64 feet; or, it is equal
to f of the rectangle A B C E.
SECTION XIV.— TRIGONOMETRY.
473. — Right- Angled Triangle§: The Sides. — In right-
angled triangles, when two sides are given, the third side
may be found by the relation of equality which exists of
the squares of the sides (Arts. 353 and 416). For example,
if the sides a and b (Fig. 313) are given, cy the third side,
may be computed from equation (115.) —
Extracting the square root, we have —
When the hypothenuse and one side are given, by transposi
tion of the factors in (115.), we have —
(A.)
or —
TWO SIDES GIVEN TO FIND THE THIRD. 511
Owing to the factors being involved to the second power in
this expression, the labor of computation is greater than
that in a more simple method, which will now be shown.
In equation (A.) or (B.) the factors under the radical may
be simplified. By equation (i 14.) we have —
Therefore, equation (A.) becomes—
a = \(c + £) (V _ fy
a form easy of solution.
For example : let c equal 29-732 and b equal 13-216, then
we have —
29-732
13-216
The sum = 42-948
The difference = 16-516
By the use of a table of logarithms (Art. 427) the problem
may be easily solved ; thus —
Log. 42-948 = 1-6329429
16-516 = i -2179049
To get the square root— 2)2- 8508478
a — 26-6332 = i -4254239
This method is applicable to the sides of a triangle, only ;
for the hypothenuse it will not serve. The length of the
hypothenuse as well as that of either side may, however, be
obtained by proportion ; provided a triangle of known di-
mensions and with like angles be also given.
For example: in Fig. 314, in which the two sides a and
b are known, let it be required to find c, the hypothenuse.
Draw the line D E parallel with A C, then the two trian-
gles B DE and BAG are homologous; consequently their
512
TRIGONOMETRY.
corresponding sides are in proportion (Art. 361). Hence, if
d equals unity, we have —
d : f : : a : c,
= */,'
from which, when a and / are known, c is obtained by sim-
ple multiplication.
474. — Right- Angled Triangle§: Trigonometrical Ta-
bles— To render the simple method last named available,
the lengths of d, e and f (Fig. 314) have been computed for
triangles of all possible angles, and the results arranged in
FIG. 314.
tables, termed Trigonometrical Tables. The lines d, e, and
/, are known as sines, cosine, tangents, cotangents, etc., as
shown in Fig. 315 — wheVe A B is the radius of the circle
B C H. Draw a line A F, from A, through any point, C, of the
arc B G. From C draw CD perpendicular to A B ; from B
draw BE perpendicular to A B ; and from G draw G F per-
pendicular to A G.
Then, lor the angle FA B, when the radius A C equals
unity, CD is the sine; AD the cosine; DB the versed sine ;
BE the tangent; GF the cotangent ; AE the secant; and
A Fihe cosecant.
But if the angle be larger than one right angle, yet less
than two right angles, as BAH, extend HA to K and E B
to K, and from H draw H J perpendicular to A J.
TRIGONOMETRICAL TABLES.
513
Then, for the angle BAH, when the radius A H equals
unity, HJ is the sine ; A J the cosine; BJ the versed sine ;
B K the tangent ; and A K the secant.
When the number of degrees contained in a given angle
is known, the value of the sine, cosine, etc., corresponding to
that angle, may be found in a table of Natural Sines, CO-
FIG. 315.
sines, etc. Or, the logarithms of the sines, cosines, etc., may
be found in logarithmic tables.
In the absence of such a table, and when the degrees
contained in the given angle are unknown, the values of
the sine, cosine, etc., may be found by computation, as fol-
lows:— Let ABC (Fig. 316) be the given angle. At any
distance from B draw b perpendicular to B C. By any scale
of equal parts obtain the length of each of the three lines a,
b, c. Then for the angle at B we have, by proportion —
5M-
TRIGONOMETRY.
c : b : : i -o : sin. B = — .
c
c : a : : I «o : cos. B = — .
c
a : b : : i -o : tan. B = —.
a
b : a : : I -o : cot. B — -.
a i c : : i -o : sec. B = -.
a
b : c : : i • o : cosec. B — --.
Or, in any right-angled triangle, for the angle contained
between the base and hypothenuse —
When perp. divided by hyp., the quotient equals the sine.
base " " hyp., " " " cosine.
" perp. " " base, " " tangent.
" base " " perp., " " cotangent.
" hyp. " " base, " " " secant.
" hyp. " " perp., " " cosecant.
To designate the angle to which a trigonometrical term
applies, the letter at the intended angle is annexed to the
c
FIG. 316.
name of the trigonometrical term ; thus, in the above exam-
ple, for the sine of A B C we write sin. B ; for the cosine,
cos. B, etc.
EQUATIONS TO RIGHT-ANGLED TRIANGLES.
515
By these proportions the two acute angles of a right-
angled triangle may be computed, provided two of the
sides are known. For when the perpendicular and hypoth-
enuse are known, the sine and cosecant may be obtained.
When the base and hypothenuse are known, the cosine and
secant may be computed. And when the base and perpen-
dicular are known, the tangent and cotangent may be com-
puted.
Either one of these, thus obtained, shows by the trigo-
nometrical tables the number of degrees in the angle ; and,
deducting the angle thus found from 90°, the remainder will
be the angle of the other acute angle of the triangle. For
M D
FIG. 317.
example : in a right-angled triangle, of which the base is 8
feet and the perpendicular 6 feet, how many degrees are
contained in each of the acute angles ?
Having, in this case, the base and perpendicular known,
by referring to the above proportions we find that with
these two sides we may obtain the tangent ; therefore—
Referring to the trigonometrical tables, we find that 0-75 is
the tangent of 36° 52' 12", nearly ; therefore—
The quadrant equals 90- o- o
The angle B equals 36-52-12
The angle A equals 53-07-48
TRIGONOMETRY.
475. — Right-Angled Triangle§ : Trigonometrical Value
of Side§.— In the triangle A B C (Fig. 317), with BP = i for
radius, and on B as a centre, describe the arc P D, and from
its intersection with the lines A B and B C, draw PM and
T D perpendicular to the line B C. Then from homologous
triangles we have these proportions for the perpendicular —
BD : DT : : BC : CA,
r : tan. B : : base : perp.,
I : tan. B : : a : b = a tan. B. (1%S-)
Also— *
- BP : PM \\BA\AC,
r : sin. B : : hyp. : perp.,
I : sin. B : : c : b = c sin. B. (186.)
For the base, we have —
BP : BM : : BA : B C,
r : cos. B : : hyp. : base,
I : cos. B : : c : a = c cos. B. (l%7-)
Again—
TD : BD :: AC : B C,
tan. B : r : : perp. : base,
tan.*:!::*:,^ L_. (.88.)
For the hypothenuse, we have —
PM : PB : : A C : : A B,
sin. B : r : : perp. : hyp.,
RULE FOR THE PERPENDICULAR. 517
sin. B : i : : b : c = -, . (180.)
sm. B
Again —
BD \BT\\BC\ BA,
r : sec. B : : base : hyp.,
I : sec. B : : a : c = a sec. B — — - — . (IQO.)
cos. B
This substitution of the cos. for the sec. is needed because
tables of secants are not always accessible. That it is an
equivalent is clear ; for we have —
BM \ BP :-. BD : BT,
cos. \r\\r\ sec. =
cos.
By these equations either side of a right-angled triangle
may be computed, provided there are certain parts of the
triangle given. As, for example : of the six parts of a tri-
angle (the three sides and the three angles), three must be
given, and at least one of these must be a side.
As an example : let it be required to find two sides of a
right-angled triangle of which the base is 100 feet, and the
acute angle at the base is 35 degrees. Here we have given
one side and two angles (the base, acute angle, and the right
angle) to find the other two sides, the perpendicular and the
hypothenuse.
Among the above rules we have, in equation (185.), for
the perpendicular —
b — a tan. B.
Or : The perpendicular equals the product of the base into the
tangent of the acute angle at the base.
5 1 8 TRIGONOMETRY.
Then (Art. 427)—
The logarithmic tangent of B (— 35°) is 9-8452268
Log. of a (— 100) is 2-0000000
Perpendicular, b (= 70-02075) = I -8452268
And for the hypothenuse, taking equation (190.), we
have —
cos. B
Or : The hypothenuse equals the quotient of the base divided
by the cosine of the acute angle at the base.
For this we have-
Log, of a (= 100) is 2-0000000
" cos. B (= 35°) is 9-9133645
Hypothenuse c(~ 122-0775) = 2-0866355
We thus find that a right-angled triangle, having an angle
of 35 degrees at the base, has its three sides, the perpendic-
ular, baseband hypothenuse, respectively equal to 70-02075,
loo, and 122-0775.
N.B. — The angle at A (Fig^ij) is obtained by deducting
the angle at B from 90° (Art. 346). Thus, 90 — 35 = 55 ;
this is the angle at A, in the above case.
If the perpendicular be given, then for the base use
equation (188.), and for the hypothenuse use equation (189.).
If the hypothenuse be given, then for the base use equation
(187.) and for the perpendicular use equation (186.).
USEFUL RULE FOR THE SIDES. 519
476. — Oblique-Angled Triangle* : Sine* and Side*. — In
the oblique-angled triangle A BC (Fig. 318) from C and per-
pendicular to A B draw CD. This line divides the oblique-
angled triangle into two right-angled triangles, the lines and
angles of which may be treated by the rules already given ;
but there is a still more simple method, as will now be
shown.
As shown in Art. ^4: " When the perpendicular is di-
vided by the hypothenuse the quotient equals the sine."
Applying this to Fig. 318, we have —
jNIVERSITY
Let the former be divided by the latter ; then —
d
sin. A __ b_
~sm7£ ~ d '
a
or, reducing, we have —
sin. A
or, putting the equation in the form of a proportion-
sin. B : sin. A : : b : a ;
or ; the sines are in proportion as the sides, respectively op-
posite. Or, as commonly stated, the sines are in proportion
as the sides which subtend them.
This is a rule of great utility ; by it we obtain the follow-
ing :
Referring to Fig. 318, we have-
sin. B : sin. A : : b : a = b *-. — - . (191.)
520 TRIGONOMETRY.
~ sin. A
sin. C : sin. A : : c : a = c- ~ . (192.)
sin. C
„ , sin. B
sin. A : sin. B : : a : b — a —. . (ICH.)
sin. A
~ . „ , sin. B
sin. C : sin. B : : c : b = £- -. (iQ4.)
sin. (7
, . ^ sin. 6" , N
sin. A : sin. c : : a : c = a-^— -~ . (iQ5.)
sin. A
•n ' r L 7 Sm- C
sin. B : sin. C \ \ b \ c — b .
sin. B
These expressions give the values of the three sides respec-
tively ; two expressions for each, one for each of the two
remaining sides; that is to be used which contains the given
side.
From these expressions we derive the values of the
sines \ thus —
sin. A = sin. Ba-. (197-)
b
sin. A = sin. C — . (19%-)
sin. B = sin. A —. (I99«)
• ci
sin. B = sin. C '-. (200.)
sin. C — sin. A -. (201.)
sin. C — sin. BC-. (202.)
477. — Oblique - Angled Triangles : First Class. — The
problems arising in the treatment of oblique-angled trian-
gles have been divided into four classes, one of which, the
TO FIND THE TWO SIDES. 521
first, will here be referred to. The problems of the first
class are those in which a side and two angles are given, to
find the remaining angle and sides.
As to the required angle, since the three angles of every
triangle amount to just two right angles (Art. 345), or 180°,
the third angle may be found simply by deducting the sum
of the two given angles from 180°.
For example : referring to Fig. 318, if angle A = 18° and
angle B = 42°, then their sum is 18 + 42 = 60, and 180 -
60 = 120° = the angle AC B.
To find the two sides : if a be the given side, then to find
the side b we have, equation (193.)—
sin. B
b — a ~ ;
sin. A
or, the side b equals the product of the side a into the quo-
tient obtained by a division of the sine of the angle opposite
b by the sine of the angle opposite a.
For example: in a triangle (Fig. 318) in which the angle
A = 1 8°, the angle B — 42° (and, consequently (Art. 345) the
angle C= 120°), and the given side a equals 43 feet; what
are the lengths of the sides b and cl Equation (193.) gives-
sin.' B
b~a- --.
sin. A
Performing the problem by logarithms (Art. 427), we
have —
Log. a(= 43)= I -6334685
Sin. B (- 42°) = 9-8255109
£-4589794
Sin. A (= 1 8°) = 9-4899824
Log. b O 93 « 1 102) = i -9689970.
Thus the side b equals 93-1102 feet, or 93 feet I inch and
nearly one third of ah inch.
522 TRIGONOMETRY.
For the side c, we have, equation (195.) —
sin. C
c = a —. — — ;
sin. A
or —
Log. *(= 43) =1-6334685
Sin. C(= 120°) = 9-9375306
1' 5 709991
Sin. A (— 1 8°) = 9-4899824
Log. c(= 120-508) — 2-0810167
or, the base c equals 120 feet 6 inches and one tenth of an
inch, nearly. But if instead of a the side b be given, then
for a use equation (191.), and for c use equation (196.).
And, lastly, if c be the given side, then for a use equation
(192.), and for b use equation (194.). t
478. — Oblique-Angled Triangles: Second Clas§.— The
problems which comprise the second class are those in which
.two sides ^ax\& an angle opposite to one of them are given, to
find the two remaining angles and the third side.
The only requirement really needed here is to find a
second angle ; for, with this second angle found, the problem
is reduced to one of the fir§t class ; and the third side may
then be found under rules given in Art. 477.
To find a second angle, use one of the equations (197.) to
(202.).
For example : in the triangle ABC (Fig. 318), let a (= 43)
and b (= 93 • 1 1) be the two given sides, and A, the angle op-
posite a, be the given angle (= 18°). Then to find the angle
B, we have equation (199.) — (selecting that which in the
right hand member contains the given angle and sides) —
sin. B = sin. A —
43
OBLIQUE-ANGLED TRIANGLES. 523
By logarithms (Art. 427), we have —
Log. sin. A (= 1 8°) = 9-4899824
" 93-n = 1.9689970
1-4589794
43 = 1-6334685
" sin. 5 (=42°) = 9-8255109
By reference to the log. tables, the last line of figures, as
above, is found to be the sine of 42° ; therefore, the required
angle B is 42°. Then 180° - (18° + 42°) = 120° = the angle C.
With these angles, or with any two of them, the third
side c may be found by rules given in Art. 477.
FIG. 319.
479. — Oblique-Angled Triangles : Sum and Difference
of Two Angles. — Preliminary to a consideration of prob-
lems in the third class of triangles, it is requisite to show the
relation between the sum and difference of two angles.
In Fig. 319, let the angle A JM and the angle A JN be
the two given angles ; and let A J M be called angle A, and
AJN, angle B. Now the sum and difference of the angles
may be ascertained by the use of the sum and difference of
the sines of the angles, and by the sum and difference of the
tangents. In the diagram, in which the radius A J equals
524 TRIGONOMETRY.
unity, we have MP, the sine of angle A (= A y M), and
NQ = RP, the sine of angle B(= A y N). Then—
MP- RP= MR
equals the difference of the sines of the angles ; and since
PM' = PM—
PM'+RP = RM',
equals the sum of the sines of the angles.
With the radius JC describe the arc J D E, and tangent
to this arc draw FH parallel with MM', or perpendicular
to AB.
Then FD is the tangent of the angle M C N, and D H is
the tangent of the angle NCM'.
Now since an angle at the circumference is equal to half
the angle at the centre standing on the same arc (Art. 355),
therefore the measure of the angle M C N is the half of M N,
equals—
-B).
Similarly, we have —
for the angle NCM'.
Therefore we have for the tangent of the angle M C N
-B\
and, for the tangent of the angle N C M'-
DH = tan. \(A + B).
And, because FC D and M C R are homologous triangles, as,
also, DCH and R'CM', therefore—
• M' R : MR : : DH : D F,
SUM AND DIFFERENCE OF TWO ANGLES. 525
sin. A 4- sin. B : sin. A — sin. B : : tan. %(A + B) : tan. ±(A — B),
from which we have —
sin. A — sin. B __ tan. % (A — B) ,~ .
sin. ^4 + sin. ./? . tan. % (A + B)'
To obtain a proper substitute for the first member of this
expression we have, equation (195.)—
sin. C
/• 7>
C- —* LI ; ~— .
sin. A
or —
c sin. A — a sin. C. (M.)
We also have, equation (196.) —
,sin. C
c = b -. ,
sin. B
or —
c sin. B — b sin. C. (N.)
These two equations, (M.) and (N.), added, give —
c sin. A -t- c sin. B — a sin. (7 + b sin. C
or —
c (sin. ^ + sin. B) = sin. C(« + b). (P.)
But, if equation (N.) be subtracted from equation (M.), we
have —
c sin. A — c sin. B — a sin. C — b sin. C,
or —
f(sin. A — sin. B) = A sin. C (a — £).' (R.)
If equation (R.) be divided by equation (P.), we have—
<r(sin. A — sin. B) __ sin. C(a — b)
<r(sin. A -f sin. B) ~ sin. C (a + b) '
526 TRIGONOMETRY.
which reduces to —
sin. A — sin. B _ a — b
sin. A + sin. B ~ a + b'
The first member of this equation is identical with the first
member of the above equation (D.), and therefore its equal,
the second member, may be substituted for it ; thus—
a — b _ tan, j- (A - B)
a +~b~ tan. ^(A~+^)"
From which we have — t
tan. $(A—£) = tan. J (A + B) ~ — . (203.)
We have (Art. 431) the proposition, that if half the differ-
ence of two quantities be subtracted from half their sum, the
remainder will equal the smaller quantity. For example :
if A represent the larger quantity and B the smaller, then —
)«> Jfj (204.)
and, again, we also have (Art. 431) —
B)=A. (205.)
480.— Oblique-Angled Triangles; : Third < In**. — The
third class of problems comprises all those cases in which two
sides of a triangle and their included angle are given, to
find the other side and angles.
In this case, as in the problems of the second class, the
only requirement here is to find a second angle ; for then
the problem becomes one belonging to the first class. But
the finding of the second angle, in problems of the third
class, is attended with more computation than it is in prob-
lems of the second class. The process is as follows : Hav-
ing one angle of a triangle, the sum of the two remaining
OBLIQUE-ANGLED TRIANGLES. 527
angles is obtained by subtracting the given angle from
1 80° — the sum of the three angles.
Then with equation (203.) the difference of the two angles
is obtained. And then, having the sum and difference of the
two angles, either may be found by one of the equations
(204.) and (205.).
For example : let Fig. 320 represent the triangle in
which a (= 36 feet) and b(= 27 feet) are the given sides ; and
C (= 105°) the angle included between the given sides, a and
b. The sum of the two angles A and B, therefore, will be —
(A + £) = 180- 105 ^75°,
and the half of the sum of A and B is -V - 37° 30'.
The sum of the given sides is 36 + 27 = 63, and their dif-
ference is 36 — 27 = 9.
Then from equation (203.) we have—
. tan. ±(A-B) = tan. 37
Solving this by logs. (Art. 427), we have-
Log. tan. 37° 30' = 9-8849805
9 =0-9542425
0-8392230
63 = I-79934Q5
tan. $(A- B)(= 6° 15' 20-5") = 9-0398825
Thus half the difference of A and B is 6° 15' 20- 5", nearly,
528 TRIGONOMETRY.
By equation (204.) —
37° 30'
6° 15' 20- 5"
The difference, 31° 14' 39'$"
and by equation (205.)—
37-30
6.15.20-5
The sum, 43.45.20-5 — A
From above, 31.14.39-5 — ^
The given angle, 105. o. o — C
The three angles, 180. o. o
Thus, by adding together the three angles, the work is
tested and proved.
Having the three angles, the third side may now be
found by the rule for problems of the first class.
— Oblique- Angled Triangles: Fourth Class. — The
fourth class comprises those problems in which the three
sides of the triangle are given, to find the three angles.
The method by which the problems of the fourth class
are solved is to divide the triangle into two right-angled
triangles; then, by the use of equation (129.), to find one
side of one of these triangles, and then with this side to find
one of the angles, then by rules for the second class prob-
lems, obtain the second and third angles.
Thus, from equation (129.), we have —
By the relation of sines to sides (Art. 476), we have (Fig.
b : g : : sin. E : sin. F.
TRIANGLES— FOURTH CLASS.
529
But the angle E is a right angle, of which the sine is unity,
therefore^ —
b : g : : I : sin. F = -.
Substituting for g its value as above, we have —
*sin. F =
q- b)
2bc
(206.)
To illustrate: let a, b, c (Fig. 321) be the three given sides
of the triangle ABC, respectively equal to 12,' 8 and 16 feet.
With these, equation (206.) becomes —
I62-(i2 + 8)(i2-8)
sin. F = *-*»-< —^ ,
2 x 8 x 16
sin. /* =
sin. F
176
Solving this by logarithms (Art. 427), we have-
Log. 176 = 2-2455127
" 256 = 2-4082400
Log. sin. 43° 26' = 9*8372727
or, the angle at F equals 43° 26', nearly. Of the triangle
530
TRIGONOMETRY.
A C E (Fig. 321), E is a right angle, therefore the sum of F
and A, the two remaining angles, equals 90° (Art. 346).
Hence, for the angle at A, we have —
=90° -43° 26' = 46° 34'-
We now have two sides a and b and A, an angle opposite
to one of them, to find B, a second angle. For this, equa-
tion (199.) is appropriate. Thus —
sin. B = sin. A —.
a
This may be solved as shown in Art. 478.
And, when the second angle is obtained, the third angle
is found by subtracting the sum of the first and second an-
gles from 1 80°.
But to test the accuracy of the work, it is well to com-
pute the angle 'C from the angle A, and the sides a and c.
For this, equation (201.) will be appropriate.
482.— Trigonometric Formulae: Right- Angled Trian-
gles.— For facility of reference the formulas of previous
articles are here presented in tabular form. The symbols
referred to are those of Fig. 322.
FORMULA IN TABULAR FORM.
531
RIGHT-ANGLED TRIANGLES.
GIVEN.
REQUIRED.
FORMULA.
a, b,
a, c,
b, c,
;
c = fV+J*.
£ = V(c + *) (c
-a).
a = V(c + ^) (^
-b).
A,
A,
£= 90° -A.
A = 90° — ^.
B, a,
*'
^ = « tan. ^.
rt:
" cos. B '
B,b,
*,
b
tan. ^ *
b
B.,.
a,
b,
a = c cos. B,
b — c sin. B.
483. Trigonometrical Formulae: First Class, Oblique.
C
—The symbols of the formulae of the following- table indi-
cate quantities represented in Fig. 323 by like symbols.
532
TRIGONOMETRY.
OBLIQUE-ANGLED TRIANGLES: FIRST CLASS.
GIVEN, i REQUIRED.
FORMULA.
;
A "D /~>
•fJ. y Jj) O.
^=180 — ^+^.
A, C, B,
B = iSo-A + C.
B, C, A,
A= 180 - B + C.
A B b \ a
..sin. A
n.t jjf t/y \ u-,
[
A C c \ a
sin. B'
sin. A
JJ.) W', C-, | C*,
' . W- <s ^-, •
sm. (7
1
A B // h
, sin. ^
A /7
f*9 JJj l*j \ I/,
1
B C c b
sm. ^4
, sin. B
h — r
U) C-, t, C/j
sm. C
A C /r r
sin. £7
si, L,, a, cy
\
sin. A
,sin. £7
'•*''. '
t — u • . Ty.
sm. j5
484. — Trigonometrical Formulae : Second Cla§§, Oblique.
—The symbols in the formulae of the following table refer
to quantities represented in Fig. 323, by like symbols.
FORMULAE FOR TRIANGLES, SECOND CLASS. 533
OBLIQUE-ANGLED TRIANGLES: SECOND CLASS.
1
GIVEN.
REQUIRED
FORMULAE.
B, a, b,
A,
sin. A = sin. B —.
b
C, a, c,
A,
sin. A — sin. C — .
c
A, a, b,
By
sin. B = sin. A —.
a
C, b, c,
By
sin. B = sin. C — .
c
A, a, c,
c,
sin. C = sin. A — .
a
B, b, c,
Cy
sin. C — sin. B-.
B, C,
A,
A — 180 — B + C.
A, C,
Ay By
By
B = 1 80 — A + C.
C = 1 80 — A + B.
For —
ay
See Formulae, First Class.
534
TRIGONOMETRY.
485. — Trigonometrical Formulae : Third Class, Oblique.
— The symbols in the formulae of the following table refer
to quantities shown by like symbols in Fig. 323.
OBLIQUE-ANGLED TRIANGLES: THIRD CLASS.
GIVEN.
REQUIRED.
FORMULA.
C, a, b,
A +B,
A- B,
A,
tan. $(A-B) = tan. $(A+ B) ~-^.
B,
B =$(A+B) — i(A — B).
A, b, c,
C +B,
C-B,
C + B = 1 80 — A.
c — b
tan. 4- (C — B) = tan. £ (C + B) — -—. .
c + b
r —— 1 ( f JL. 7?'\ _J_ 1 /" f 7?\
D 1 ( i^ i /?\ 1 /'/^ J2\
-D —— "2" Ivx T *'l — 2 \^ — **1*
B, a, c,
C + A.
r A
C + A — 180 — B.
£ ft
U SI,
c,
c + a
A,
A=t(C+A) + i(C-A).
For the remaining side consult formulas for the first
class.
486. — Trigonometrical Formulae: Fourtli Class, Ob-
lique.— The symbols in the formulae of the following table
refer to quantities shown by like symbols in Fig. 321.
FORMULA FOR TRIANGLES, FOURTH CLASS. 535
OBLIQUE-ANGLED TRIANGLES: FOURTH CLASS.
Given «, #, c, to find A, B, C.
A = 90 - F.
sin. ^ = sin. A — .
^ /i c
sin. c = sin. ^4 -.
= 180 — (A + B}.
SECTION XV.— DRAWING.
487. — General Remark§. — A knowledge of the proper-
ties and principles of lines can best be acquired by practice.
Although the various diagrams throughout this work may
be understood by inspection, yet they will be impressed
upon the mind with much greater force, if they are actually
drawn out with pencil and paper by the student. Science
is acquired by study — art by practice ; he, therefore, who
would have anything more than a theoretical (which must
of necessity be a superficial) knowledge of carpentry and
geometry, will provide himself with the articles here speci-
fied, and perform all the operations described in the fore-
going and following pages. Many of the problems may
appear, at the first reading, somewhat confused and intricate ;
but by making one line at a time, according to the explana-
tions, the student will not only succeed in copying the fig-
ures correctly, but by ordinary attention will learn the
principles upon which they are based, and thus be able to
make them available in any unexpected case to which they
may apply.
488.— Articles Required. — The following articles are
necessary for drawing, viz. : a drawing-board, paper, draw-
ing-pins or mouth-glue, a sponge, a T-square, a set-square,
two straight-edges, or flat rulers, a lead pencil, a piece of
india-rubber, a cake of india-ink, a set of drawing-instru-
ments, and a scale of equal parts.
489. — The Drawing-Board. — The size of the drawing-
board must be regulated according to the size of the draw-
ings which are to be made upon it. Yet for ordinary prac-
tice, in learning to draw, a board about fifteen by twenty
inches, and one inch thick, will be found large enough, and
DRAWING PAPER. 537
more convenient than a larger one. This board should be
well seasoned, perfectly square at the corners, and without
clamps on the ends. A board is better without clamps,
because the little service they are supposed to render by
preventing the board from warping is overbalanced by the
consideration that the shrinking of the panel leaves the
ends of the clamps projecting beyond the edge of the board,
and thus interfering with the proper working of the stock
of the T-square. When the stuff is well-seasoned, the warp-
ing of the board will be but trifling ; and by exposing the
rounding side to the fire, or to the sun, it may be brought
back to its proper shape.
490. — Drawing-Paper. — For mere line drawings, it is
unnecessary to use the best drawing-paper ; and since, where
much is used, the expense will be considerable, it is desirable
for economy to procure a paper of as low a price as will be
suitable for the purpose. The best paper is made in Eng-
land and water-marked " Whatman." This is a hand-made
paper. There is also a machine-made paper at about half-
price, and the manilla paper, of various tints of russet color,
is still less in price. These papers are of the various sAzes
needed, and are quite sufficient for ordinary drawings.
49 1. — To Secure the Paper to the Board. — A drawing-
pin is a small brass button, having a steel pin projecting from
the underside. By having one of these at each corner, the
paper can be fixed to the board ; but this can be done in a
better manner with moutJi-glue. The pins will prevent the
paper from changing its position on the board ; but, more
than this,, the glue keeps the paper perfectly tight and
smooth, thus making it so much the more pleasant to work
on.
To attach the paper with mouth-glue, lay it with the
bottom side up, on the board ; and with a straight-edge and
penknife cut off the rough and uneven edge. With a
sponge moderately wet rub all the surface* of the paper,
except a strip around the edge about half an inch wide. As
soon as the glistening of the water disappears turn the sheet
DRAWING. •
over and place it upon the board just where you wish it
glued. Commence upon one of the longest sides, and pro-
ceed thus : lay a flat ruler upon the paper, parallel to the
edge, and within a quarter of an inch of it. With a knife,
or anything similar, turn up. the edge of the paper against
the edge of the ruler, and put one end of the cake of mouth-
glue between your lips to dampen it. Then holding it up-
right, rub it against and along the entire edge of the paper
that is turned up against the ruler, bearing moderately
against the edge of the ruler, which must be held firmly
with the left hand. Moisten the glue as often as it becomes
dry, until a sufficiency of it is rubbed on the edge of the
paper. -Take away the ruler, restore the turned-up edge to
the level of the board, and lay upon it a strip of pretty stiff
paper. By rubbing upon this, not very hard but pretty
rapidly, with the thumb-nail of the right hand, so as to cause
a gentle friction and heat to be imparted to the glue that is
on the edge of the paper, you will make it adhere to the
board. The other edges in succession must be treated in
the same manner.
Some short distances along one or more of the edges
may afterward be found loose ; if so, the glue must again
be applied, and the paper rubbed until it adheres. The
board must then be laid away in a warm or dry place ; and
in a short, time the surface of the paper will be drawn out,
perfectly tight and smooth, and ready for use. The paper
dries best when the board is laid level. When the drawing
is finished lay a straight-edge upon the paper and cut it
from the board, leaving the glued strip still attached. This
may afterward be taken off by wetting it freely with the
sponge, which will soak the glue and loosen the paper. Do
this as soon as the drawing is taken off, in order that the
board may be dry when it is wanted for use again. Care
must be taken that, in applying the glue, the edge of the
paper does not become damper than the rest ; if it should,
the paper must be laid aside to dry (to use at another time)
and another sheet be used in its place.
Sometimes, especially when the draAving-board is new,
the paper will not stick very readily ; but by persevering
THE T-SQUARE. 539
this difficulty may be overcome. In the place of the mouth-
glue a strong solution of gum-arabic may be used, and on
some accounts is to be preferred ; for the edges of the paper
need not be kept dry, and it adheres more readily. Dissolve
the gum in a sufficiency of warm water to make it of the
consistency of linseed-oil. It must be applied to the paper
with a brush, when the edge is turned up against the ruler,
as was described for the mouth-glue. If two drawing-boards
are used, one may be in use while the other is laid away to
dry; and as' they may be cheaply made, it is advisable to
have two. The drawing-board having a frame around it,
commonly called a panel board, may afford rather more
facility in attaching the paper when this is of the size to
FIG. 324.
suit ; yet it has objections which overbalance that consid-
eration.
492. — The T-Square. — A T-square of mahogany, at once
simple in its construction and affording all necessary service,
may be thus made : let the stock or handle be seven inches
long, two and a quarter inches wide, and three eighths of an
inch thick ; the blade, twenty inches long (exclusive of the
stock), two inches wide, and one eighth of an inch thick. In
joining the blade to the stock, a very firm and simple joint
may be made by dovetailing it — as shown at Fig. 324.
493. — Tlie Set-Square. — The set-square is in the form of
a right-angled triangle ; and is commonly made of mahogany,
540 DRAWING.
one eighth of an inch in thickness. The size that is most
convenient for general use is six inches and three inches
respectively for the sides which contain the right angle,
although a particular length for the sides is by no means
necessary. Care should be taken to have the square corner
exactly true. This, as also the T-square and rulers, should
have a hole bored through them, by which to hang them
upon a nail when not in use.
494. — The Rulers. — One of the rulers may be about
twenty inches long, and the other six inches. The pencil
ought to be hard enough to retain a fine point, and yet not
so hard as to leave ineffaceable marks. It should be used
lightly, so that the extra marks that are not needed when
the drawing is inked, may be easily rubbed off with the
rubber. The best kind of india-ink is that which will easily
rub off upon the plate ; and, when the cake is rubbed against
the teeth, will be free from grit.
495. — The Instruments. — The drawing-instruments may
be purchased of mathematical instrument makers at various
prices ; from one to one hundred dollars a set. In choosing
a set, remember that the lowest price articles are not always
the cheapest. A set, comprising a. sufficient number of
instruments for ordinary use, well made and fitted in a ma-
hogany box, may be purchased of the mathematical instru-
ment makers in New York for four or five dollars. But for
permanent use those which come at ten or twelve dollars
will be found to be better.
496. — The Scale of Equal Parts. — The best scale of
equal parts for carpenters' use, is one that has one eighth,
three sixteenths, one fourth, three eighths, one half, five
eighths, three fourths, and seven eighths of an inch, arid one
inch, severally divided into twelfths, instead of being divided,
as they usually are, into tenths. By this, if it be required
to proportion a drawing so that every foot of the object
represented will upon the paper measure one fourth of an
inch*, use that part of the scale which is divided into one
THE SET-SQUARE. 541
fourths of an inch, taking for every foot one of those divis-
ions, and for every inch one of the subdivisions into twelfths ;
and proceed in like manner in proportioning a drawing to
any of the other divisions of the scale. An instrument in
the form of a semi-circle, called a protractor, and used for
laying down and measuring angles, is of much service to
surveyors, and occasionally to carpenters.
497. — The U§e of the Set-Square. — In drawing parallel
lines, when they are to be parallel to either side of the
board, use the T-square ; but when it is required to draw
lines parallel to a line which is drawn in a direction oblique
FIG. 325.
to either side of the board, the set-square must be used.
Let ab (Fig. 325) be a line, parallel to which it is desired to
draw one or more lines. Place any edge, as c d, of the set-
square even with said line ; then place the ruler gh against
one of the other sides, as ce, and hold it firmly ; slide the
set-square along the edge of the ruler as far as it is desired,
as at /; and a line drawn by the edge *'/ will be parallel
to a b.
To draw a line, as kl (Fig. 326), perpendicular to another,
as a b, set the shortest edge of the set-square at the line a b ;
place the ruler against the longest side (the hypothenuse of
the right-angled triangle); hold the ruler firmly, and slide
the set-square along until the side ed touches the point k\
then the line Ik, drawn by it, will be perpendicular to ab.
542 DRAWING.
In like manner, the drawing of other problems may be facil-
itated, as will be discovered in using the instruments.
498. — Directions for Drawing. — In drawing a problem,
proceed, with the pencil sharpened to a point, to lay down
the several lines until the whole figure is completed, ob-
serving to let the lines cross each other at the several angles,
instead of merely meeting. By this, the length of every
line will be clearly defined. With a drop or two of water,
rub one end of the cake of ink upon a plate or saucer, until
a sufficiency adheres to it. Be careful to dry the cake of
FIG 326.
ink ; because if it is left wet it will crack and crumble in
pieces. With an inferior camel's-hair pencil add a little
water to the ink that was rubbed on the plate, and mix it
well. It should be diluted sufficiently to flow freely from
the pen, and yet be thick enough to make a black line. With
the hair pencil place a little of the ink between the nibs of
the drawing-pen, and screw the nibs together until the pen
makes a fine line. Beginning with the curved lines, proceed
to ink all the lines of the figure, being careful now to make
every line of its requisite length. If tl.ey are a trifle too
short or too long the drawing will have a ragged appear-
ance ; and this is opposed to that neatness and accuracy
which is indispensable -to a good drawing. When the ink
is dry efface the pencil-marks with the india-rubber. If the
PUTTING THE DRAWING IN INK. 543
pencil is used lightly they will all rub off, leaving those lines
only that were inked.
In problems all auxiliary lines are drawn light ; while the
lines given and those sought, in order to be distinguished at
a glance, are made much heavier. The heavy lines are
made so by passing over them a second time, having the
nibs of the pen separated far enough to make the lines as
heavy as desired. If the heavy lines are made before the
drawing is cleaned with the rubber they will not appear so
black and neat, because the india-rubber takes away part
of the ink. If the drawing is a ground-plan or elevation of
a house, the shade-lines, as they are termed, should not be
put in until the drawing is shaded ; as there is danger of the
heavy lines spreading when the brush, in shading or color-
ing, passes over them. If the lines are inked with common
writing-ink they will, however fine they may be made, be
subject to the same evil ; for which reason india-ink is the
only kind to be used.
SECTION XVI.— PRACTICAL GEOMETRY.
499. — Definitions. — Geometry treats of the properties of
magnitudes.
A point has neither length, breadth, nor thickness.
A line has length only.
Superficies has length and breadth only.
A plane is a surface, perfectly straight and even in every
direction ; as the face of a panel when not warped nor
winding.
A solid has length, breadth, and thickness.
A right, or straight, line is the shortest that can be drawn
between two points.
Parallel lines are equidistant throughout their length.
FIG. 327. FIG. 328. FIG. 329.
An angle is the inclination of two lines towards one an-
other (Fig. 327).
A right angle has one line perpendicular to the other
(Fig. 328).
An oblique angle is either greater or less than a right
angle (Figs. 327 and 329).
An acute angle is less than a right angle (Fig. 327).
An obtuse angle is greater than a right angle (Fig. 329).
When an angle is denoted by three letters, the middle
one, in the order they stand, denotes the angular point, and
the other two the sides containing the angle ; thus, let a, b, c
(Fig. 327) be the angle, then b will be the angular point, and
ab and be will be the two sides containing that angle.
TRIANGLES AND RECTANGLES.
545
A triangle -is a superficies having three sides and angles
(Figs. 330, 331, 332, and 333).
An equilateral triangle has its three sides equal (Fig. 330).
An isosceles triangle has only two sides equal (Fig. 331).
FIG. 330.
FIG 331.
A scalene triangle has all its sides unequal (Fig. 332).
A right-angled triangle has one right angle (Fig. 333).
An acute-angled triangle has all its angles acute (Figs. 330
and 331).
FIG. 332.
FIG. 333.
An obtuse-angled triangle has one obtuse angle (Fig. 332).
A quadrangle has four sides and four angles (Figs. 334 to
339).
A parallelogram is a quadrangle having its opposite sides
parallel (Figs. 334 to 337).
FIG. 334-
FIG. 335.
A rectangle is a parallelogram, its angles being right
angles (Figs. 334 and 335).
A square is a rectangle having equal sides (Fig. 334).
A rhombus is an equilateral parallelogram having oblique
angles (Fig. 336).
546 PRACTICAL GEOMETRY.
A rhomboid is a parallelogram having oblique angles
(Fig- 337).
A trapezoid is a quadrangle having only two of its sides
parallel (Fig. 338).
FIG. 336. FIG. 337.
A trapezium is a quadrangle which has no two of its sides
parallel (Fig. 339).
A polygon is a figure bounded by right lines.
A regular polygon has its sides and angles equal.
An irregular polygon has its sides and angles unequal.
FIG. 338. FIG. 339.
A trigon is a polygon of three sides (Figs. 330 to 333) ; a
tetragon has four sides (Figs. 334 to 339) ; a pentagon has five
(Fig. 340) ; a hexagon six (Fig. 341) ; a heptagon seven (/%-.
342) ; an octagon eight (/r/£~. 343) ; a nonagon nine ; a decagon
ten ; an undecagon eleven ; and a dodecagon twelve sides.
FIG. 340. FIG. 341. FIG. 342. FIG. 343.
A circle is a figure bounded by a curved line, called the
circumference, which is everywhere equidistant from a cer-
tain point within, called its centre.
The circumference is also called the periphery, and some-
times the circle.
PARTS OF THE CIRCLE.
547
The radius of a circle is a right line drawn from the
centre to any point in the circumference (ab, Fig. 334).
All the radii of a circle are equal.
The diameter is a right line passing through the centre,
and terminating at two opposite points in the circumference.
Hence it is twice the length of the radius (cd. Fig. 344.)
FIG. 344.
An arc of a circle is a part of the circumference (cb, or
bed, Fig. 344).
A chord is a right line joining the extremities of an arc
(b d, Fig. 344).
A segment is any part of a circle bounded by an arc and
its chord (A, Fig. 344).
FIG. 345.
A sector is any part of a circle bounded by an arc and
two radii, drawn to its extremities (B, Fig. 344).
A quadrant, or quarter of a circle, is a sector having a
quarter of the circumference for its arc (C, Fig. 344).
A tangent is a right line which, in passing a curve,
touches, withont cutting it (fg, Fig. 344).
PRACTICAL GEOMETRY.
A cone is a solid figure standing upon a circular base di-
minishing in straight lines to a point at the top, called its
vertex (Fig: 345).
The axis of a cone is a right line passing through It,
from the vertex to the centre of the circle at the base.
An ellipsis is described if a cone be cut by a plane, not
parallel to its base, passing quite through the curved surface
(a b, Fig. 346).
A parabola is described if a cone be cut by a plane, par-
allel to a plane touching the curved surface (c d, Fig. 346 —
cd being parallel to f g\
An hyperbola is described if a cone be cut by a plane,
FIG. 347. t
parallel to any plane within the cone that passes through its
vertex (e/t, Fig. 346).
Foci are the points at which the pins are placed in de-
scribing an ellipse (see Art. 548, and /,/, Fig. 347).
The transverse axis is the longest diameter of the ellipsis
(a b, Fig. 347).
The conjugate axis is the shortest diameter of the ellipsis ;
and is, therefore, at right angles to the transverse axis (cd,
Fig. 347).
The parameter is a right line passing through the focus
of an ellipsis, at right angles to the transverse axis, and ter-
minated by the curve (gk and gt, Fig. 347).
RIGHT LINES AND ANGLES.
549
A diameter of an ellipsis is any right line passing through
the centre, and terminated by the curve (kl, or ;;/ n, Fig. 347).
A diameter is conjugate to another when it is parallel to a
tangent drawn at the extremity of that other — thus, the di-
ameter mn (Fig. 347) being parallel to the tangent op, is
therefore conjugate to the diameter kl.
A double ordinate is any right line, crossing a diameter of
an ellipsis, and drawn parallel to a tangent at the extremity
of that diameter (it, Fig. 347).
A cylinder is a solid generated by the% revolution of a
right-angled parallelogram, or rectangle, about one of its
FIG. 348.
FIG. 349.
sides ; and consequently the ends of the cylinder are equal
circles (Fig. 348).
The axis of a cylinder is a right line passing through it
from the centres of the two circles which form the ends.
A segment of a cylinder is comprehended under three
planes, and the curved surface of the cylinder. Two of
these are segments of circles ; the other plane is a parallelo-
gram, called by way of distinction, the plane of tJie segment.
The circular segments are called the ends of the cylinder
(Fig. 349)-
PROBLEMS.
RIGHT LINES AND ANGLES.
500. — To Bi§ect a Line. — Upon the ends of the line ab
(Fig. 350) as centres, with any distance for radius greater
than half ab, describe arcs cutting each other in
550
PRACTICAL GEOMETRY.
draw the line cd, and the point e, where it cuts ab} will be
the middle of the line ab.
In practice, a line is generally divided with the com-
passes, or dividers ; but this problem is useful where it is
desired to draw, at the middle of another line, one at right
angles to it. (See Art. 514.)
501. — To Erect a Perpendicular. — From the point a
(Fig. 351) set off any distance, as ab, and the same distance
from a to c ; upon c, as a centre, with any distance for radius
greater than ca, describe an arc at d\ upon b, with the same
FIG. 351.
radius, describe another at d\ join d and a, and the line da
will be the perpendicular required.
This, and the three following problems, are more easily
performed by the use of the set-square (see Art. 493). Yet
they are useful when the operation is so large that a set-
square cannot be used.
TO ERECT A PERPENDICULAR.
551
502. — To let Fall a Perpendicular. — Let a (Fig. 352) be
the point above the line be from which the perpendicular is
required to fall. Upon a, with any radius greater than ad,
describe an arc, cutting be at ^and /; upon the points e and
/, with any radius greater than ed, describe arcs, cutting
FIG. 352.
each other at g\ join a and g, and the line ad will be the
perpendicular required.
503. — To Erect a Perpendicular at the End of a Line.
—Let a (Fig. 353), at the end of the line c a, be the point at
which the perpendicular is to be erected. Take any point,
as by above the line ca, and with the radius ba describe the
arc dae; through d and b draw the line de\ join e and a,
then e a will be the perpendicular required.
FIG. 353-
The principle here made use of is a very important one,
and is applied in many other cases (see Art. 510, 3d, and Art.
513. For proof of its correctness, see Art. 352).
A second method. Let b (Fig. 354), at the end of the line
a b, be the point at which it is required to erect a perpendic-
ular. Upon b, with any radius less than b a, describe the arc
ced\ upon ct with the same radius, describe the small arc at*/
552
PRACTICAL GEOMETRY.
and upon c, another at d ; upon e and d, with the same or any
other radius greater than half e d, describe arcs intersecting
at f' join /and b, and the line fb will be the perpendicular
required. This method of erecting a perpendicular, and
that of the following article, depend for accuracy upon the
c b
FIG. 354.
fact that the side of a hexagon is equal to the radius of the
circumscribing circle.
A third method. Let b (Fig. 355) be the given point at
which it is required to erect a perpendicular. Upon b, with
any radius less than ba, describe the quadrant def\ upon d,
with the same radius, describe an arc at e, and upon e an-
other at c ; through d and e draw dc, cutting the arc in c ;
join c and 3, then cb will be the perpendicular required.
d b
FIG. 355.
This problem can be solved by the six, eight and ten rule,
as it is called, which is founded upon the same principle as
the problems at Arts. 536, 537, and is applied as follows:
let ad (Fig. 353) equal eight, and ae, six ; then, if de equals
ten, the angle cad is a right angle. Because the square of
six and that of eight, added together, equal the square of
EQUAL ANGLES. 553
ten, thus : 6 x 6 = 36, and S x 8 — 64; 36 + 64 = 100, and
10 x 10 = 100. Any sizes, taken in the same proportion, as
six, eight and ten, will produce the same effect ; as 3, 4 and
5, or 12, 1 6 and 20. (See Art. 536.)
By the process shown at Fig. 353, the end of a board may
be squared without a carpenters'-square. All that is neces-
sary is a pair of compasses and a ruler. Let ca be the edge
of the board, and a the point at which it is required to be
squared. Take the point b as near as possible at an angle
of forty-five degrees, or on a mitre-line from a, and at about
the middle of the board. This is not necessary to the work-
ing of the problem, nor does it affect its accuracy, but the
result is more easily obtained. Stretch the compasses from
b to a, and then bring the leg at a around to d\ draw a line
from d, through b, out indefinitely ; take the distance db and
place it from b to c ; join e and a ; then ca will be at right
angles to c a. In squaring the foundation of a building, or
laying out a garden, a rod and chalk-line may be used in-
stead of compasses and ruler.
504. — To let Fall a Perpendicular near the End of a
Line. — Let e (Fig. 353) be the point above the line c a, from
which the perpendicular is required to fall. From e draw
any line, as e d, obliquely to the line ca; bisect edai b\ upon
b, with the radius be, describe the arc ead\ join e and #;
then ea will be the perpendicular required.
505. — To Make an Angle (a§ edf, Fig. 356*) Equal to a
Given Angle (as b a c).— From the angular point a, with any
FIG. 356.
radius, describe the arc b c ; and with the same radius, on
the line dc, and from the point d, describe the arc/^-; take
the distance be, and upon gt describe the small arc at/;
554
PRACTICAL GEOMETRY.
join f and d\ and the angle edf will be equal to the angle
bac.
If the given line upon which the angle is to be made is
situated parallel to the. similar line of the given angle, this
may be performed more readily with the set-square. (See
Art. 497.)
506. — To Bisect an Angle. — Let a be (Fig. 357) be the
angle to be bisected. Upon b, with any radius, describe the
FIG. 357-
arc a c ; upon a and c, with a radius greater than half a c,
describe arcs cutting each other at d; join b and d\ and bd
will bisect the angle a be, as was required.
This problem is frequently made use of in solving other
problems ; it should therefore be well impressed upon the
memory.
507 — To Trisect a Right Angle.— Upon a (Fig. 358),
with any radius, describe the arc b c ; upon b and c, with the
FIG. 358.
same radius, describe arcs cutting the arc be at d and e\
from d and e draw lines to a, and they will trisect the angle,
as was required.
TO DIVIDE A GIVEN LINE
The truth of this is made evident by the following oper-
ation : divide a circle into quadrants ; also, take the radius
in the dividers, and space off the circumference. This will
divide the circumference into just six parts. A semi-circum-
ference, therefore, is equal to three, and a quadrant to one
and a half of those parts. The radius, therefore, is equal to
two thirds of a quadrant ; and this is equal to a right angle.
508. — Through a Given Point, to Draw a Line Parallel
to a Given Line. — Let a (Fig. 359) be the given point, and
FIG. 359.
be the given line. Upon any point, as d, in the line be, with
the radius da, describe the arc ac\ upon a, with the same
radius, describe the arc de\ make de equal to ac\ through
e and a draw the line ea, which will be the line required.
This is upon the same principle as Art. 505.
509. — To Divide a Given Line into any Number of
Equal Part§. — Let a b (Fig. 360) be the given line, and 5 the
number of parts. Draw ac at any angle to a b ; on ac, from
J A'
FIG. 360.
a, set off five equal parts of any length, as at i, 2, 3, 4 and c ;
join c and b\ through the points I, 2, 3, and 4, draw I e , 2/,
3^ and 4//, parallel to cb\ which will divide the line ab, as
was required.
556
PRACTICAL GEOMETRY.
The lines ab and ac are divided in the same proportion.
(See Art. 542.)
THE CIRCLE.
510. — To Find the Centre of a Circle. — Draw any chord,
as ab (Fig. 361), and bisect it with the perpendicular cd\ bi-
sect cd with the line e f, as at g\ then g is the centre, as was
required.
A second method. Upon any two points in the circumfer-
ence nearly opposite, as a and b (Fig. 362), describe arcs cut-
ting each other at c and d; take aay other two points, as e
and fy and describe arcs intersecting, as at g and h ; join g
and h and c and d\ the intersection o is the centre.
This is upon the same principle as Art. 514.
A third method. Draw any chord, as ab (Fig. 363), and
from the point a draw ac at right angles to ab ; join c and
b\ bisect c b at d — which will be the centre of the circle.
A TANGENT AT A GIVEN POINT.
557
If a circle be not too large for the purpose, its centre
may very readily be ascertained by the help of a carpenters'-
square, thus : apply the corner of the square to any point in
the circumference, as at a ; by the edges of the square
(which the lines ab and ac represent) draw lines cutting the
FIG. 363.
circle, as at b and c ; join b and c ; then if be is bisected, as at
d, the point d will be the centre. (See Art. 352.)
5 II „ — At a Given Point in a Circle to Draw a Tangent
thereto. — Let a (Fig. 364) be the given point, and b the cen-
FIG. 364.
tre of the circle. Join a and b ; through the point a, and at
right angles to a b, draw cd ; then c d is the tangent required.
512. — The Same, without making use of the Centre of
the Circle. — Let a (Fig. 365) be the given point. From a set
off any distance to b, and the same from b to c ; join a and
c ; upon a, with ab for radius, describe the arc dbc-, make
db equal to bc\ through a and d draw a line; this will be
the tangent required.
558 PRACTICAL GEOMETRY.
The correctness of this method depends upon the fact
that the angle formed by a chord and tangent is equal to any
inscribed angle in the opposite segment of the circle (Art.
358); ab being the chord, and bca the angle in the opposite
segment of the circle. Now, the angles dab and bca are
equal, because the angles dab and bac are, by construction,
FIG. 365.
equal; and the angles bac and bca are equal, because the
triangle abc is an isosceles triangle, having its two sides, ab
and be, by construction equal ; therefore the angles dab and
bca are equal.
513. — A Circle and a Tangent Given, to Find Hie Point
of Contact.— From any point, as a (Fig. 366), in the tangent
FIG. 366.
be, draw a line to the centre d\ bisect ad at r; upon e, with
the radius ea, describe the arc afd\ f is the point of con-
tact required.
If/ and// were joined, the line would form right angles
with the tangent be. (See Art. 352.)
A CIRCLE THROUGH GIVEN POINTS.
559
514. — Through any Three Points not in a Straight Line,
to Draw a Circle. — Let a, b and c (Fig. 367) be the three
given points. Upon a and b, with any radius greater than
half a b, describe arcs intersecting at d and e ; upon b and c,
with any radius greater than half be, describe arcs intersect-
ing at / and g\ through d and e draw a right line, also
.. FIG. 367.
another through / and g\ upon the intersection //, with the
radius ha, describe the circle a be, and it will be the one re-
quired.
515. — Three Points not in a Straight Line being Given,
to Find a Fourth that shall, with the Three, Lie in the
Circumference of a Circle. — Let a b c (Fig. 368) be the given
points. Connect them with right lines, forming the triangle
FIG. 368.
acb', bisect the angle cba (Art. 506) with the line bd\ also
bisect c a in e, and erect ed perpendicular to ac, cutting bd
in d; then d is the fourth point required.
A fifth point may be found, as at /, by assuming a, d and
b, as the three given points, and proceeding as before. So,
560 PRACTICAL GEOMETRY.
also, any number of points may be found simply by using
any three already found. This problem will be serviceable
in obtaining short pieces of very flat sweeps. (See Art. 240.)
The proof of the correctness of this method is found in
the fact that equal chords subtend equal angles (Art. 357).
Join d and c; then since ae and ec are, by construction,
equal, therefore the chords a d and dc are equal ; hence the
angles they subtend, dba and d b c, are equal. So, like-
wise, chords drawn from a to /, and from / to d, are equal,
and subtend the equal angles dbf and fba. Additional
points beyond a or b may be obtained on the same principle.
To obtain a point beyond a, on b, as a centre, describe with
any radius the arc ion ; make on equal to o i ; through b and
n draw b g\ on a as centre and with af for radius, describe
the arc, cutting gb at gt then g. is the point sought.
516. — To Describe a Segment of a Circle toy a Set-Tri-
angle.— Let a b (Fig. 369) be the chord, and c d the height
FIG. 369.
of the segment. Secure two straight-edges, or rulers, in the
position ce and c f, by nailing them together at c, and affixing
a brace from c to /; put in pins at a and b ; move the angu-
lar point c in the direction acb\ keeping the edges of the
triangle hard against the pins a and b ; a pencil held at c
will describe the arc acb.
A curve described by this process is accurately circular,
and is not a mere approximation to a circular arc, as some
may suppose. This method produces a circular curve, be-
cause all inscribed angles on one side of a chord-line are
equal (Art. 356). To obtain the radius from a chord and its
versed sine, see Art. 444.
If the angle formed by the rulers at c be a right angle,
TO FIND THE VERSED SINE. 561
the segment described will be a semi-circle. This problem
is useful in describing centres for brick arches, when they
are required to be rather flat. Also, for the head hang-
ing-stile of a window-frame, where a brick arch, instead of a
stone lintel, is to be placed over it.
517. — To Find the Radiu§ of an Arc of a Circle when
the Chord and Ver§ed Sine are Given. — The radius is equal
to the sum of the squares of half the chord and of the versed
sine, divided by twice the versed sine. This is expressed,
(-}* a
algebraically, thus : r = — — , where r is the radius, c the
chord, and v the versed sine (Art. 444).
Example. — In a given arc of a circle a chord of 12 feet
has the rise at the middle, or the versed sine, equal to 2 feet,
what is the radius ?
Half the chord equals 6, the square of 6 is, 6 x 6 = 36
The square of the versed sine is, 2x2=4
Their sum equals, 40
Twice the versed sine equals 4, and 40 divided by 4 equals
10. Therefore the radius, in this case, is 10 feet. This
result is shown in less space and more neatly by using the
above algebraical formula. For the letters substituting
their value, the formula r = — — i- — becomes r = sj£
2V 2X2
and performing the arithmetical operations here indicated
equals —
6 a -f 2 2 _ 36 -t- 4 _ 40 _
4 44"
518. — To Find the Ver§ed Sine of an Arc of a Circle
when the Radius and Chord are Given. — The versed sine
is equal to the radius, less the square root of the difference
of the squares of the radius and half chord ; expressed alge-
braically thus : v = r — Vr 3 - (l)a, where r is the radius, v
the versed sine, and c the chord. (Equation (161.) reduced.)
562 PRACTICAL GEOMETRY.
Example. — In an arc of a circle whose radius is 75 feet,
what is the versed sine to a chord of 120 feet? By the table
in the Appendix it will be seen that —
The square of the radius, 75, equals . . 5625
The square of half the chord, 60, equals . 5600
The .difference is ..... 2025
The square root of this is . . . -45
This deducted from the radius ... 75
•
The remainder is the versed sine, = 30
This is expressed by the formula, thus —
v = 75 - ^75 " - FF? = 75 - ^5625-3600 = 75 - 45 = 30.
519. — To Describe the Segment of a Circle by Intersec-
tion of Lines. — Let ab (Fig. 370) be the chord, and cd the
height of the segment. Through c draw ef parallel to a b ;
draw bf at right angles to cb; make ce equal to cf; draw
ag and bh at right angles to a b ; divide ce, cft da, db, a g,
and bh, each into a like number of equal parts, as four;
draw the lines i 1,22, etc., and from the points o, o, and o,
draw lines to c\ at the intersection of these lines trace the
curve, acb, which will be the segment required.
In very large work, or in laying out ornamental gar-
dens, etc., this will be found useful ; and where the centre
of the proposed arc of a circle is inaccessible it will be inval-
uable. (To trace the curve, see note at Art. 550.)
The lines e a, c d, and fb, would, were they extended,
meet in a point, and that point would be in the opposite
side of the circumference of the circle of which acb is a
ORDINATES TO AN ARC.
563
segment. The lines i i, 2 2, 3 3, would likewise, if extended,
meet in the same point. The line cd, if extended to the op-
posite side of the circle, would become a diameter. The line
fb forms, by construction, a right angle with be, and hence
the extension of fb would also form a right angle with be,
on the opposite side of bc\ and this right angle would be
the inscribed angle in the semi-circle ; and since this is re-
quired to be a right angle (Art. 352), therefore the construc-
tion thus far is correct, and it will be found likewise that at
each point in the curve formed by the intersection of the
radiating lines, these intersecting lines are at right angles.
520. — Ordinate§. — Points in the circumference of a
circle may be obtained arithmetically, and positively accu-
rate, by the calculation of ordinates, or the parallel lines o i,
« -f J 2 / (1725
FIG. 371.
02, 03, 04 (Fig. 3/i). These ordinates are drawn at right
angles to the chord-line a b, and they may be drawn at any
distance apart, either equally distant or unequally, and there
may be as many of them as is desirable ; the more there are
the more points in the curve will be obtained. If they are
located in pairs, equally distant from the versed sine c d,
calculation need be made only for those on one side of cd,
as those on the opposite side will be of equal lengths, re-
spectively; for example: o i, on the left-hand side of cd, is
equal to o i on the right-hand side, o 2 on the right equals
o 2 on the left, and in like manner for the others.
The length of any ordinate is equal to the square root
of the difference of the squares of the radius and abscissa,
less the difference between the radius and versed sine (Art.
445). The abscissa being the distance from the foot of
the versed sine to the foot of the ordinate. Algebraically,
564 PRACTICAL GEOMETRY.
t = Vr * •- x* — (r — b\ where t is put to represent the ordi-
nate ; x, the abscissa ; b, the versed sine ; and r, the radius.
Example. — An arc of a circle has its chord ab (Fig. 371)
100 feet long, and its versed sine cd, 5 feet. It is required
to ascertain the length of ordinates for a sufficient number
of points through which to describe the curve. To this end
it is requisite, first, to ascertain the radius. This is readily
/£\2 2
done in accordance with Art. 517. For - becomes
- = 252-5 = radius. Having the radius, the curve
2x5
might at once be described without the ordinate points, but
for the impracticability that usually occurs, in large, flat
segments of the circle, of getting a location for the centre,
the centre usually being inaccessible. The ordinates are,
therefore, to be calculated. In Fig. 371 the ordinates are
located equidistant, and are 10 feet apart. It will only
be requisite, therefore, to calculate those on one side of
the versed sine cd. For the first ordinate 01, the formula
/ = Vr^— x* — (r — &) becomes—
2- io2-(252-5 -5).
= 1/63756-25 — 100 — 247.5.
252.3019-247.5.
4.8019 = the first ordinate, o i.
For the second —
t— ^252- 52 — 202 — (252-5 — 5).
= 251-7066 — 247.5.
4-2066 = the second ordinate, 02.
For the third—
/ = 1/^52-5 a-3o2 - 247. 5.
= 250-7115-247.5.
3-2115 = the third ordinate, 03.
TO DESCRIBE A TANGED CURVE. 565
For the fourth—
/ = 1/252. 5 2 -402 - 247- 5.
= 249-3115 - 247.5.
1-8115 = the fourth ordinate, o 4.
The results here obtained are in feet and decimals of a
foot. To reduce these to feet, inches, and eighths of an
inch, proceed as at Reduction of Decimals in the Appendix.
If the two-feet rule, used by carpenters and others, were
decimally divided, there would be no necessity of this re-
duction, and it is to be hoped that the rule will yet be thus
divided, as such a reform would much lessen the labor of
computations, and insure more accurate measurements.
Versed sine c d = ft. 5 «o = ft. 5 -o inches.
Ordinates o I = 4-8019= 4-9! inches, nearly.
" 02= 4-2066 = 4. 2\ inches, nearly.
" 03= 3-2115 — 3 -2j inches, nearly.
" 04= 1-8115 = I- 9f inches, nearly.
521. — In a Given Angle, to Describe a Tanged Curve.
— Let a b c (Fig. 372) be the given angle, and I in the line a b,
FIG. 372.
and 5 in the line be, the termination of the curve. Divide
i b and b 5 into a like number of equal parts, as at i, 2, 3, 4,
and 5 ; join i and i, 2 and 2, 3 and 3, etc. ; and a regular
curve wiU be formed that will be tangical to the line ab, at
the point i, and to be at 5.
This is of much use in stair-building, in easing the angles
formed between the wall-string and the base of the hall,
also between the front string and level facia, and in many
other instances. The curve is not circular, but of the form
of the parabola (Fig. 418) ; yet in large angles the difference
PRACTICAL GEOMETRY.
is not perceptible. This problem can be applied to describ-
ing- the curve for door-heads, window-heads, etc., to rather
better advantage .than Art. 516. For instance, let ab (Fig.
373) be the width of the opening, and c d the height of the
« c b
FIG. 373.
arc. Extend c d, and make de equal to cd\ join a and e,
also e and b ; and proceed as directed above.
522. — To De§eribe a Circle within any Given Triangle,
so that the Sides of the Triangle shall be Tangical. —
Let a be (Fig. 374) be the given triangle. Bisect the angles
FIG. 374.
a and b according to Art. 506; upon d, the point of intersec-
tion of the bisecting lines, with the radius d e, describe the
required circle.
523. — About a Given Circle, to Describe an Equilateral
Triangle. — Let adb c (Fig. 375) be the given circle. Draw
the diameter c d; upon'af, with the radius of the given circle,
describe the arc aeb\ join a and b ; draw fg at right angles
to dc ; make fc and eg each equal to ab ; from ft through
a, draw //*, also from g, through b, draw gh\ then fgh
will be the triangle required.
524. — To Find a Right L.ine nearly Equal to the Cir-
cumference of a Circle. — Let abed (Fig. 376) be the given
A RIGHT LINE EQUAL TO A CIRCUMFERENCE. 567
circle. Draw the diameter ac\ on this erect an equilateral
triangle aec according to Art. 525 ; draw gf parallel to ac\
extend ec to /, also ea to g\ then gf will be nearly the
* ^ FIG. 375.
length of the semi-circle adc\ and twice gf will nearly
equal the circumference of the circle a b c d, as was required.
Lines drawn from *, through any points in the circle, as
o, o and o, to /,/ and /, will divide gf in the same way as
the semi-circle adc is divided. So, any portion of a circle
may be transferred to a straight line. This is a very useful
ff P P d p f
FIG. 376.
problem, and should be well studied, as it is frequently
used to solve problems on stairs, domes, etc.
Another method. Let a bfc (Fig. 377) be the given circle.
Draw the diameter ac\ from d, the centre, and at right an-
568 PRACTICAL GEOMETRY.
gles to ac, draw db\ join b and c\ bisect be at e\ from d,
through e, draw df\ then ef added to three times the di-
ameter, will equal the circumference of the circle sufficiently
near for many uses. The result is a trifle too large. If the
FIG. 377.
circumference found by this rule be divided by 648-22 the
quotient will be the excess. Deduct this excess, and the
remainder will be the true circumference. This problem is
rather more curious than useful, as it is less labor to perform
the operation arithmetically, simply multiplying the given
diameter by 3- 1416, or, where a greater degree of accuracy
is needed, by 3-1415926. (See Art. 446.)
POLYGONS, ETC.
525. — Upon a Given Line to Construct an Equilateral
Triangle. — Let a b (Fig. 378) be the given line. Upon a and
«
FIG. 378.
b, with a b for radius, describe arcs, intersecting at c ; join a
and c, also c and b ; then acb will be the triangle required.
526. — To De§cribe an Equilateral Rectangle, or Square.
— Let a b (Fig. 379) be the length of a side of the proposed
POLYGONS IN CIRCUMSCRIBING CIRCLES.
square. Upon a and b, with a b for radius, describe the arcs
a d and b c ; bisect the arc ae in f.\ upon e, with ef for ra-
dius, describe the arc cfd\ join a and <:, c and </, </and £;
then acdb will be the square required.
FIG. 379.
527. — Within a Given Circle, to Inscribe an Equilateral
Triangle, Hexagon or Dodecagon. — Let abed (Fig. 380) be
the given circle. Draw the diameter bd\ upon b, with the
radius of the given circle, describe the arc ae c ; join a and c,
also a and </, and c and </— and the triangle is completed.
For the hexagon: from a, also from c, through e, draw the
lines af and cg\ join # and b, b and <:, c and /, etc., and the
hexagon is completed. The dodecagon may be formed by
bisecting the sides of the hexagon.
Each side of a regular hexagon is exactly equal to the
radius of the circle that circumscribes the figure. For the
radius is equal to a chord of an arc of 60 degrees ; and, as
every circle is supposed to be divided into 360 degrees, there
is just 6 times 60, or 6 arcs of 60 degrees, in the whole cir-
cumference. A line drawn from each angle of the hexagon
570
PRACTICAL GEOMETRY.
to the centre (as in the figure) divides it into six equal, equi-
lateral triangles.
528. — Within a Square to Inscribe an Octagon. — Let
abed (Fig. 381) be the given square. Draw the diagonals
a d and b c ; upon a, b, c, and d, with a e for radius, describe
arcs cutting the sides of the square at 1,2, 3, 4, 5, 6, 7, and 8 ;
join i and 2, 3 and 4, 5 and 6, etc., and the figure is com-
pleted.
In order to eight-square a hand-rail, or any piece that is
FIG. 382.
to be afterwards rounded, draw the diagonals a d and b c
upon the end of it, after it has been squared-up. Set a
gauge to the distance ae and run it upon the whole length
of the stuff, from each corner both ways. This will show
BUTTRESSED OCTAGON. 5/1
how much is to be chamfered off, in order to make the piece
octagonal (Art. 354).
529. — To Find the Side of a Buttressed Octagon. — Let
ABCDE (Fig. 382) represent one quarter of an octagon
structure, having a buttress H FGJ at each angle. The
distance M H, between the buttresses, being given, as also
F G, the width of a buttress ; to find H C or C J, in order to
obtain B C, the side of the octagon. Let B C, a side of the
octagon, be represented by b ; or D C by £ b. Let M H = a ;
or J D = ±a\ and
Then we have —
JD+ JC = CD,
. i a + x = i b,
a + 2 x = b.
For FG put p\ ^r LG — K J — \p.
Now £ D is the radius of an inscribed circle and, as per
equation (140.), equals r — ( 1/2 + i) -.
Also, £ (7 is the radius of a circumscribed circle, and, as
per equation (141.)* equals R— ^2^2 + 4-.
The two triangles, CJK and C ED, are homologous;
for the angles at C are common and the angles at AT and D
are right angles. Having thus two angles of one equal
respectively to the two angles of the other, therefore (Art.
345) the remaining angles must be equal. Hence, the sides
of the triangles are proportionate, or—
ED : EC :: JK : CJ
r • R '.'. \}
The value of the side, as above, is—
R
I = a + 2 x = a + p - - ,
572 pkACTICAL GEOMETRY.
And taking the value of R and r, as above, we have —
(1/2
+1
r>
Substituting this for — , we have —
V2 + I
The numerical coefficient of / reduces to 1-0823923 or
i -0824, nearly.
Therefore we have —
b = a + i -o824/. (207.)
Or: The side of a buttressed octagon equals the distance be-
tween the buttresses plus \ -0824 times the width of the faced
the buttress.
For example : let there be an octagon building, which
measures between the buttresses, as at M H, 18 feet, and the
face of the buttresses, as FG, equals 3 feet ; what, in such a
building, is the length of a side B Cl For this, using equa-
tion (207.), we have —
b = 1 8 + I -0824 x 3
= 18 + 3-2472
= 21-2472.
Or : The side of the octagon B C equals 21 feet and nearly 3
inches.
530. — Within a Given Circle to In§cribe any Regular
Polygon. — Let abc 2 (Figs. 383* 384, and 385) be given circles.
Draw the diameter a c ; upon this erect an equilateral trian-
gle aec, according to A rt. 525 ; divide ac into as many equal
parts as the polygon is to have sides, as at i, 2, 3, 4, etc.;
from e, through each even number, as 2, 4, 6, etc., draw lines
TO DESCRIBE ANY REGULAR POLYGON.
573
cutting the circle in the points 2, 4, etc.; from these points
and at right angles to a c draw lines to the opposite part
of the circle ; this will give the remaining points for the
polygon, as b, /, etc.
In forming a hexagon, the sides of the triangle erected
FIG. 383.
upon ac (as at Fig. 384) mark the points b and f. This
method of locating the angles of a polygon is an approxima-
tion sufficiently near for many purposes ; it is based upon
the like principle with the method of obtaining a right line
nearly equal to a circle (Art. 524). The method shown at
Art. 531 is accurate.
FIG. 386.
FIG. 387.-
FIG. 388
531.— Upon a Given Line to De§crifoe any Regular
Polygon.— Let a b (Figs. 386, 387, and 388) be (riven lines,
equal to a side of the required figure. From b draw be at
right angles to a b ; upon a and &, with a b for radius, describe
the arcs acd and feb\ divide ac into as many equal parts
5/4 PRACTICAL GEOMETRY.
as the polygon is to have sides, and extend those divisions
from c towards d\ from the second point of division, count-
ing from c towards a, as 3 (Fig. 386), 4 (Fig. 387), and 5 (Fig.
388), draw a line to b ; take the distance from said point of
division to a, and set it from b to e ; join e and a ; upon the
intersection o with the 'radius oa, describe the circle afdb;
then radiating lines, drawn from b through the even numbers
on the arc a d, will cut the circle at the several angles of the
required figure.
In the hexagon (Fig. 387), the divisions on the arc ad are
not necessary ; for the point o is at the intersection of the
arcs ad and fb, the points f and d are determined by the
intersection of those arcs with the circle, and the points
above g and h can be found by drawing lines from a and b
through the centre o. In polygons of a greater number of
sides than the hexagon the intersection o comes above the
arcs ; in such case, therefore, the lines a e and b 5 (Fig. 388)
have to be extended before they will intersect. This method
of describing polygons is founded on correct principles, and
is therefore accurate. In the circle equal arcs subtend
equal angles (Arts. 357 and 515). Although this method is
accurate, yet polygons may be described as accurately and
more simply in the following manner. It will be observed
that much of the process in this method is for the purpose
of ascertaining the centre of a circle that will circumscribe
the proposed polygon. By reference to the Table of Poly-
gons in Art. 442 it will be seen ho-w this centre may be ob-
tained arithmetically. This is the rule : multiply the given
side by the tabular radius for polygons of a like number of
sides with the proposed figure, and the product will be the
radius of the required circumscribing circle. Divide this
circle into as many equal parts as the polygon is to have
sides, connect the points of division by straight lines, and
the figure is complete. For example : It is desired to de-
scribe a polygon of 7 sides, and 20 inches a side. The tabu-
lar radius is 1-15238. This multiplied by 20, the product,
23-0476 is Ihe required radius in inches. The Rules for
the Reduction of Decimals, in the Appendix, show how to
change decimals to the fractions of a foot or an inch. From
EQUAL FIGURES.
575
this, 23 -0476 is equal to 23TV inches, nearly. It is not needed
to take all the decimals in the table, three or four of them
will give a result sufficiently near for all ordinary practice.
532. — To €on§truct a Triangle whose Side§ shall be
§everally Equal to Three Given Lines. — Let a, b and c (Fig.
389) be the given lines. Draw the line de and make it equal
FIG. 389.
c\ upon e, with b for radius, describe an arc at /; upon d,
with a for radius, describe an arc intersecting the other at/;
join d and /, also f and e ; then dfe will be the triangle
required.
533 — To Construct a Figure Equal to a Given, Right-
lined Figure. — Let abed (Fig. 390) be the given figure.
Make ef (Fig. 391) equal to cd-, upon /, with da for radius,
FIG. 390.
FIG. 391.
describe an arc at^; upon r, with ca for radius, describe an
arc intersecting the other at^; join ^and e,\ upon f and g,
with db and ab for radius, describe arcs intersecting at // ;
join g and /*, also h and /; then Fig. 391 will every way
equal Fig. 390.
So, right-lined figures of any number of sides may be
copied, by first dividing them into triangles, and then pro-
576
PRACTICAL GEOMETRY.
ceeding as above. The shape of the floor of any room, or
of any piece of land, etc., may be accurately laid out by this
problem, at a scale upon paper ; and the contents in square
feet be ascertained by the next.
534. — To Make a Parallelogram equal to a Given
Triangle. — Let a be (Fig. 392) be the given triangle. From
a draw a d at right angles to b c ; bisect a d in e ; through e
f
FIG. 392.
draw fg parallel to be-, from b and c draw b f and eg ^par-
allel to de\ then bfgc will be a parallelogram containing a
surface exactly equal to that of the triangle a be.
Unless the parallelogram is required to be a rectangle,
the lines bf and eg need not be drawn parallel to d e. If a
rhomboid is desired they may be drawn at an oblique angle,
provided they be parallel to one another. To ascertain the
area of a triangle, multiply the base be by half the perpen-
d
FIG. 393.
dicular height da.
is taken for base.
In doing this it matters not which side
535. — A Parallelogram being Given, to Construct An-
other Equal to it, and Having a Side Equal to a Given Line.
—Let A (Fig. 393) be the given parallelogram, and B the
given line. Produce the sides of the parallelogram, as at
SQUARE EQUAL TO TWO OR MORE SQUARES. 577
a, b, c, and d'\ make ed equal to B ; through d draw cf par-
allel to gb-, through e draw the diagonal ca\ from a draw
af parallel to ed; then C will be equal to A. (See Art. 340.)
536. — To Make a Square Equal to two or more Given
Squares. — Let A and B (Fig. 394) be two given squares.
FIG. 394.
Place them so as to form a right angle, as at a ; Join b ai<d c ;
then the square C, formed upon the line be, wi'J1 be equal in
extent to the squares A and B added together. Again : if
a b (Fig. 395) be equal to the side of a given square, c a, placed
at right angles to a b, be the side of another given square,
and cd, placed at right angles to cb, be the side of a third
given square, then the square A, formed upon the line db>
will be equal to the three given squares. (See Art. 353.)
The usefulness and importance of this problem are pro-
verbial. To ascertain the length of braces and of rafters in
578 PRACTICAL GEOMETRY.
framing, the length of stair-strings, etc., are some of the pur-
poses to which it may be applied in carpentry. (See note
to Art. 503.) If the lengths of any two sides of a right-
angled triangle are known, that of the third can be ascer-
tained. Because the square of the hypothenuse is equal to
the united squares of the two sides that contain the right
angle.
(i.) — The two sides containing the right angle being
known, to find the hypothenuse.
Rule. — Square each given side, add the squares together,
and from the product extract the square root ; this will be
the answer.
For instance, suppose it were required to find the length
of a rafter for a house, 34 feet wide — the ridge of the roof
to be 9 feet high, above the level of the wall-plates. Then
17 feet, half of the span, is one, and 9 feet, the height, is the
other of the sides that contain the right angle. Proceed as
directed by the rule :
17 9
17 _9
119 8 1 = square of 9.
17 289 = square of 17.
289 — square of 17. 370 Product.
i ) 370 ( 19-235 + = square root of 370 ; equal 19 feet 2-J in.,
i i nearly ; which would be the required
20 )~270 length of the rafter.
9 261
382). -900
_2 ^1
3843) 13600
38465)- 207 100
192325
TO FIND THE LENGTH OF A RAFTER. 579
(By reference to the table of square roots in the Appen-
dix, the root of almost any number may be found ready
calculated ; also, to change the decimals of a foot to inches
and parts, see Rules for the Reduction of Decimals in the
Appendix.)
Again : suppose it be required, in a frame building, to
find the length of a brace having a run of three feet each
way from the point of the right angle. The length of the
sides containing the right angle will be each 3 feet ; then, as
before —
3
_3
9 = square of one side.
3 times 3 = 9 = square of the other side.
1 8 Product : the square root of which is 4 • 2426-}- ft.,
or 4 feet 2 inches and -J full.
(2.) — The hypothenuse and one side being known, to find
the other side.
Rule. — Subtract the square of the given side from the
square of the hypothenuse, and the square root of the prod-
uct will be the answer.
Suppose it were required to ascertain the greatest per-
pendicular height a roof of a given span may have, when
pieces of timber of a given length are to be used as rafters.
Let the span be 20 feet, and the rafters of 3 x 4 hemlock
joist. These come about 13 feet long. The known hy-
pothenuse, then, is 13 feet, and the known side, 10 feet —
that being half the span of the building.
13
13
39
13
169 = square of hypothenuse.
10 times 10 = 100 — square of the given side.
69 Product : the square root of which is
580 PRACTICAL GEOMETRY.
8 • 3066+ feet, or 8 feet 3 inches and | full. This will be the
greatest perpendicular height, as required. Again : suppose
that in a story of 8 feet, from floor to floor, a step-ladder is
required, the strings of which are to be of plank 12 feet
long, and it is desirable to know the greatest run such a
length of string will afford. In this case, the two given
sides are — hypothenuse 12, perpendicular 8 feet.
12 times 12 — 144 — square of hypothenuse,
8 times 8 = 64 = square of perpendicular.
80 Product : the square root of which is
8-9442+ feet, or 8 feet n inches and ^ — the answer, as re-
quired.
Many other cases might be adduced to show the utility
of this problem. A practical and ready method of ascer-
taining the length of braces, rafters, etc., when not of a great
length, is to apply a rule across the carpenters' -square.
Suppose, for the length of a rafter, the base be 12 feet and
the height 7. Apply the rule diagonally on the square, so
that it touches 12 inches from the corner on one side, and 7
inches from the corner on the other. The number of inches
on the rule which are intercepted by the sides of the square,
13!-, nearly, wilt be the length of the rafter in feet ; viz., 13
feet and. -J- of a foot. If the dimensions are large, as 30 feet
and 20, take the half of each on the sides of the square, viz.,
15 and 10 inches; then the length in inches across will be
one half the number of feet the rafter is long. This method
is just as accurate as the preceding ; but when the length of
a very long rafter is sought, it requires great care and pre-
cision to ascertain the fractions. For the least variation on
the square, or in the length taken on the rule, would make
perhaps several inches difference in the length of the rafter.
For shorter dimensions, however, the result will be true
enough.
537. — To Make a Circle Equal to two Given Circles.—
Let A and B (Fig- 396) be the given circles. In the right-
angled triangle abc make ab equal to the diameter of the
SIMILAR FIGURES.
circle B, and cb equal to the diameter of the cin
the hypothenuse a c will be the diameter of a circle C, which
will be equal in area to the two circles A and B, added
together.
FIG. 396.
Any polygonal figure, as A (Fig. 397), formed on the hy-
pothenuse of a right-angled triangle, will be equal to two
similar figures,* as B and C, formed on the two legs of the
triangle.
FIG. 397.
538. — To €on§truct a Square Equal to a Given Rect-
angle.— Let A (Fig. 398) be the given rectangle. Extend
the side ab and make be equal to be\ bisect a c in /, and
upon /, with the radius fa, describe the semi-circle agc\
extend eb till it cuts the curve in g\ then a square bghd,
formed on the line bg, will be equal in area to the rectan-
glcA.
* Similar figures are such as have their several angles respectively equal,
and their sides respectively proportionate.
582
PRACTICAL GEOMETRY.
Another method. Let A (Fig. 399) be the given rectangle.
Extend the side a b and make a d equal to a c ; bisect a d in
e\ upon e, with the radius ea, describe the semi-circle afd\
extend gb till it cuts the curve in /; join a and f\ then
FIG. 398.
the square B, formed on the line af, will be equal in area to
the rectangle A. (See Arts. 352 and 353.)
539. — To Form a Square Equal to a Given Triangle-
Let ab (Fig. 398) equal the base of the given triangle, and be
equal half its perpendicular height (see Fig. 392) ; then pro-
ceed as directed at Art. 538.
540.— Two Right Lines being Given, to Find a Third
Proportional Thereto. — Let A and B (Fig. 400) be the given
lines. Make a b equal to A ; from a draw a c at any angle
PROPORTIONATE DIVISIONS IN LINES. 583
with ab-, make ac and ad each equal to B; join c and £;
from d draw de parallel to c b ; then # e will be the third
proportional required. That is, ae bears the same propor-
tion to B as B does to A.
FIG. 400.
541. — Three Right Lines being Given, to Find a Fourth
Proportional Thereto.— Let A, B, and C (Fig. 401) be the
given lines. Make ab equal to A ; from a draw ac at any
angle with a b ; make # c equal to j9 and a e equal to (7 ; join
c and £; from e draw */ parallel to cb\ then 0/ will be the
fourth proportional required. That is, af bears the same
proportion to C as B does to A.
To apply this problem, suppose the two axes of a given
ellipsis and the longer axis of a proposed ellipsis are given.
Then, by this problem, the length of the shorter axis to the
proposed ellipsis can be found ; so that it will bear the same
proportion to the longer axis as the shorter of the given
ellipsis does to its longer. (See also Art. 559.)
542. — A Line with Certain Divisions being Given, to
Divide Another, Longer or Shorter, Given Line in the
Same Proportion. — Let A (Fig. 402) be the line to be di-
vided, and B the line with its divisions. Make a b equal to
B with all its divisions, as at i, 2, 3', etc.; from a draw ac at
any angle with a b ; make a c equal to A ; join c and b ; from
584
PRACTICAL GEOMETRY.
the points i, 2, 3, etc., draw lines parallel to cb', then these
will divide the line ac in the same proportion as B is divided
— as was required.
This problem will be found useful in proportioning the
members of a proposed cornice, in the same proportion as
those of a given cornice of another size. (See Art. 321.) So
of a pilaster, architrave, etc.
543. — Between Two Given Right Lines, to Find a
Mean Proportional. — Let A and B (Fig. 403) be the given
lines. On the line ac make ab equal to A and be equal to
B ; bisect ac in e\ upon e, with ea for radius, describe the
semi-circle adc\ at b erect b d at right angles to a c ; then
bd will be the mean proportional between A and B. That
is, ab is to bd as bd is to be. This is usually stated thus:
ab : bd : : bd : be, and since the product of the means
equals the product of the extremes, therefore, abxbe = bd*-
This is shown geometrically at Art. 538.
CONIC SECTIONS.
544. — Definitions. — If a cone, standing upon a base
that is at right angles with its axis, be cut by a plane, per-
AXIS AND BASE OF PARABOLA.
585
pendicular to its base and passing through its axis, the sec-
tion will be an isosceles triangle (as a be, Fig. 404) ; and the
base will be a semi-circle. If a cone be cut by a plane in the
direction ef the section will be an ellipsis ; if in the direction
ml, the section will be a parabola; and if in the direction
ro, an hyperbola. (See Art. 499.) If the cutting planes be
at right angles with the plane a be, then—
545. — To Find the Axe§ of the Ellipsi§: bisect ef (Fig.
404) in g\ through g draw h i parallel to ab\ bisect h i in/;
FIG. 404.
upon j, with jh for radius, describe the semi-circle hki\
from ^-draw gk.at right angles to hi\ then twice gk will
be the conjugate axis and ef the transverse.
546. — To Find the Axis and Ba§e of the Parabola. —
Let ;;/ / (Fig. 404), parallel to ac, be the direction of the cut-
ting plane. From ;;/ draw m d at right angles to a b ; then
l-m will be the axis and height, and md an ordinate and half
the base, as at Figs. 417, 418.
547. — To Find the Height, Ba§e, and Transverse Axis
of an Hyperbola. — Let o r (Fig. 404) be the direction of the
586
PRACTICAL GEOMETRY.
cutting plane. Extend or and ac till they meet at w; from
o draw op at right angles to a b ; then r o will be the height,
n r the transverse axis, and op half the base ; as at Fig. 419.
54-8. — The Axes being Given, to Find the Foci, and to
Describe an Ellipsis with a String. — Let ab (Fig. 405) and
cd be the given axes. Upon c, with a e or be for radius, de-
scribe the arc //; then / and /, the points at which the
arc cuts the transverse axis, will be the foci. At f and f
place two pins, and another at c\ tie a string about the three
pins, so as to form the triangle ffc ; remove the pin from c
and place a pencil in its stead ; keeping the string taut,
move the pencil in the direction cga\ it will then describe
the required ellipsis. The lines fg and gf show the posi-
tion of the string when the pencil arrives at g.
This method, when performed correctly, is perfectly ac-
curate ; but the string is liable to stretch, and is, therefore,
not so good to use as the trammel. In making an ellipse by
a string or twine, that kind should be used which has the
least tendency to elasticity. For this reason, a cotton cord,
such as chalk-lines are commonly made of, is not proper for
the purpose ; a linen or flaxen cord is much better.
549. — The Axes being Given, to Describe an Ellipsis
with a Trammel. — Let ab and cd (Fig. 406) be the given
axes. Place the trammel so that a line passing through the
centre ol the grooves would coincide with the axes ; make
ELLIPSE BY TRAMMEL.
587
the distance from the pencil e to the nut/ equal to half c d\
also, from the pencil e to the nut g equal to half a b ; letting
the pins under the nuts slide in the grooves, move the tram-
mel eg in the direction cbd\ then the pencil at e will de-
scribe the required ellipse.
A trammel may be constructed thus : take two straight
strips of board, and make a groove on their face, in the cen-
tre of their width ; join them together, in the middle of their
length, at right angles to one another ; as is seen at Fig. 406.
A rod is then to be prepared, having two movable nuts
made of wood, with a mortise through them of the size of
the rod, and pins under them large enough to fill the
grooves. Make a hole at one end of the rod, in which to
FIG. 406.
place a pencil. In the absence of a regular trammel a tem-
porary one may be made, which, for any short job, will an-
swer every purpose. Fasten two straight-edges at right
angles to one another. Lay them so as to coincide with the
axes of the proposed ellipse, having the angular point at the
centre. Then, in a rod having a hole for the pencil at one
end, place two brad-awls at the distances described at Art,
549. While the pencil is moved in the direction of the
curve, keep the brad-awls hard against the straight-edges,
as directed for using the trammel-rod, and one quarter of
the ellipse will be drawn. Then, by shifting the straight-
edges, the other three quarters in succession may be drawn.
If the required ellipse be not too large, a carpenters'-square
may be made use of, in place of the straight-edges.
An improved method of constructing the trammel is as
588
PRACTICAL GEOMETRY.
follows: make the sides of the grooves bevelling from the
face of the stuff, or dove-tailing instead of square. Prepare
two slips of wood, each about two inches long, which shall
be of a shape to just fill the groove when slipped in at the
end. These, instead of pins, are to be attached one to each
of the movable nuts with a screw, loose enough for the nut
to move freely about the screw as an axis. The advantage
of this contrivance is, in preventing the nuts from slipping
out oftheir places during the operation of describing the
curve.
550. — To Describe an Ellipsis by Ordiiiute*. — Let ab
and cd (Fig. 407) be given axes. With c e or ' e d for radius
describe the quadrant fgh ; divide f/i, ac, and eb, each into
a like number of equal parts, as at I, 2, and 3 ; through
these points draw ordinates parallel to cd and -fg\ take the
distance I i and place it at i /, transfer 27 to 2 /#, and 3 k to
3 n ; through the points # , n, m, /, and c, trace a curve, and
the ellipsis will be completed.
The greater the number of divisions on a, e, etc., in this
and the following problem, the more points in the curve can
be found, and the more accurate the curve can be traced.
If pins are placed in the points n, m, /, etc., and a thin slip
of wood bent around by them, the curve can be made quite
correct. This method is mostly used in tracing face-moulds
for stair hand -railing.
551. — To Describe an Ellipsis by Intersection of Lines.
— Let ab and cd (Fig. 408) be given axes. Through c, draw
ELLIPSE BY INTERSECTION OF LINES.
589
fg parallel to.ab', from a and b draw af and b g at right
angles to a b ; divide fa, gb, ae, and eb, each into a like
number of equal parts, as at i, 2, 3, and 0, <?,<?; from I, 2,
and 3, draw lines to c; through 0, 0, and 0, draw lines from d,
intersecting those drawn to c ; then a curve, traced through
the points i, i, i, will be that of an ellipsis.
Where neither trammel nor string is at hand, this, per-
haps, is the most ready method of drawing an ellipsis. The
divisions should be small, where accuracy is desirable. By
this method an ellipsis may be traced without the axes, pro-
vided that a diameter and its conjugate be given. Thus, ab
and cd (Fig. 409) are conjugate diameters: fg\s drawn par-
allel to ab, instead of being at right angles to cd\ also, fa
and g b are drawn parallel to c d, instead of being at right
angles to a b.
590
PRACTICAL GEOMETRY.
552. — To Describe an Ellipsis by Intersecting Arcs. —
Let a b and cd (Fig. 410) be given axes. Between one of the
foci, f and f, and the centre e, mark any number of points,
at random, as I, 2, and 3 ; upon f and /, with b i for radius,
describe arcs at gt g, g, and g ; upon f and /, with a I for
radius, describe arcs intersecting the others at g>g,g, and g;
then these points of intersection will be in the curve of the
ellipsis. The other points, h and i, are found in like manner,
viz.: h is found by taking b2 for one radius, and 0,2 for the
other ; i is found by taking b 3 for one radius, and a 3 for the
other, always using the foci for centres. Then by tracing a
curve through the points c, g, //, i, b, etc., the ellipse will be
completed.
This problem is founded upon the same principle as that
of the string. This is obvious, when we reflect that the
length of the string is equal to the transverse axis, added to
TO DESCRIBE AN OVAL.
591
the distance between the foci. See Fig. 405, in which cf
equals ae, the half of the transverse axis.
553. — To Describe a Figure Nearly in the Shape of an
Ellipsis, by a Pair of Compasses. — Let ab and c d (Fig. 41 1)
be given axes. From c draw c e parallel to a b ; from a draw
ae parallel to cd\ join e and d\ bisect ea in /; join / and c,
intersecting edvn. i\ bisect ic in,*?; from o draw og at right,
angles to ic, meeting cd extended to g\ join i and g, cutting
the transverse axis in r ; make hj equal to Jig, and h k equal
to//r; from j, through r and k, draw//;z and/#; also, from
g, through /£, draw gl; upon g and/, with gc for radius,
describe the arcs il and mn\ upon r and £, with ?-# for
radius, describe the arcs ;;/*and /;/ ; this will complete the
figure.
When the axes are proportioned to one another, as at 2
to 3, the extremities, c and d, of the shortest axis, will be
the centres for describing the arcs il and m n ; and the inter-
section of ed with the transverse axis will be the centre for
describing the arc m, i, etc. As the elliptic curve is contin-
ually changing its course from that of a circle, a true ellipsis
cannot be described with a pair of compasses. The above,
therefore, is only an approximation.
554. — To Draw an Oval in the Proportion Seven by
Nine. — Let cd (Fig. 412) be the given conjugate axis. Bisect
592
PRACTICAL GEOMETRY.
cdin o, and through o draw ab at right angles to cd\ bisect
co in e ; upon o, with 0^ for radius, describe the circle efgh ;
from e, through h and ft draw <?/ and ei\ also, from ^,
through h and /, draw ^£ and gl\ upon £•, with gc for
radius, describe the arc kl] upon *, with e d for radius, de-
scribe the arc ji ; upon h and /, with h k for radius, describe
the arcs jk and /*'; this will complete the figure.
This is an approximation to an ellipsis ; and perhaps no
method can be found by which* a well-shaped oval can be
drawn with greater facility. By a little variation in the
process, ovals of different proportions may be obtained. If
quarter of the transverse axis is taken for the radius of the
circle efgh, one will be drawn in the proportion five by
seven.
FIG. 414.
555. — To Draw a Tangent to an Ellipsis. — Let abed
(Fig. 413) be the given ellipsis, and d the point of contact.
Find the foci (Art. 548) / and ft and from them, through d,
draw fe and fd\ bisect the angle (Art. 506) edo with the
line sr ; then sr will be the tangent required.
TO FIND THE AXES OF AN ELLIPSE.
593
556. — An Ellipsis with a Tangent Given, to Detect the
Point of Contact. — Let a gb f (Fig. 414) be the given ellip-
sis and tangent. Through the centre e draw a b parallel to
the tangent; anywhere between e and / draw cd parallel to
a b ; bisect cd in o ; through o and e draw fg] then g will
be the point of contact required.
557. — A Diameter of an Ellipsi§ Given, to Find its
Conjugate. — Let a b (Fig. 414) be the given diameter. Find
the line fg by the last problem ; then fg will be the diam-
eter required.
558. — Any Diameter and its Conjugate being Given, to
Ascertain the Two Axes, and thence to Describe the Ellipsis.
—Let a b and cd(Fig. 415) be the given diameters, conjugate
FIG. 415.
to one another. Through c draw cf parallel to a b ; from c
draw eg at right angles to ef\ make eg equal to ah or Jib\
join g and h ; upon g, with gc for radius, describe the arc
ikcj\ upon h, with the same radius, describe the arc In •
through the intersections / and n draw n o, cutting the tan-
gent ef in o ; upon o, with ogfor radius, describe the semi-
circle e igf\ join e and g, also g and f, cutting the arc icj
in k and / ; from r, through //, draw e m, also from /, through
h, draw // ; from k and t draw kr and ts parallel to gk
594
PRACTICAL GEOMETRY.
cutting em in r, and // in s ; make h m equal to hr, and hp
equal to hs\ then r;;/ and s p will be the axes required, by
which the ellipsis may be drawn in the usual way.
559. — To Describe an Ellipsis, whose Axes shall toe
Proportionate to the Axes of a Larger or SmalDer Given
One. — Let a cbd(Fig. 416) be the given ellipsis and axes, and
FIG. 416.
ij the transverse axis of a proposed smaller one. Join a and
c\ from i draw ie parallel to ac ; make of equal to oe ; then
ef will be the conjugate axis required, and will bear the
same proportion to ij as cd does to ab. (See Art. 541.)
560. — To Describe a Parabola by Intersection of Lines.
— Let ml (Fig. 417) be the axis and height (see Fig. 404) and
i 2
/ 3 2 1
1 2 3 in 3 2
FIG. 417.
dd a double ordinate and base of the proposed parabola.
Through / draw a a parallel to dd\ through d and d draw
da and da parallel to ml\ divide ad and dm, each into a
like number of equal parts ; from each point of division in
TO DESCRIBE AN HYPERBOLA.
595
dm draw the lines i i, 22, etc., parallel to ;;//; from each
point of division in da draw lines to/; then a curve traced
through the points of intersection o, o, and o, will be that of
a parabola.
Another method. Let m I (Fig. 418) be the axis and height,
and dd the base. Extend m I and make la equal to m I ;
join a and d, and a and d\ divide ad and ad, each into a
like number of equal parts, as at i, 2, 3, etc. ; join i and i, 2
and 2, etc., and the parabola will be completed. (See Arts.
460 to 472.)
561. — To Describe an Hyperbola by Intergection of
Lines. — Let ro (Fig. 419) be the height,// the base, and nr
the transverse axis. (See Fig. 404.) Through r draw a a
m
FIG. 418.
1 2 3 o 3 2 \ p
FIG. 419.
parallel to pp\ from / draw ap parallel to ro\ divide ap
and po, each into a like number of equal parts ; from each
of the points of division in the base, draw lines to ;/ ; from
each of the points of division in ap, draw lines to r\ then
a curve traced through the points of intersection o, o, etc.,
will be that of an hyperbola.
The parabola and hyperbola afford handsome curves for
various mouldings. (See Figs. 191 to 205 ; 222 to 224; 241
and 242 ; also note to Art. 318.)
SECTION XVII.— SHADOWS.
562. — The Art of Drawing consists in representing
solids Upon a plane surface, so that a curious and nice ad-
justment of lines is made to present the same appearance to
the eye as does the human figure, a tree, or a house. It is
by the effects of light, in its reflection, shade, and shadow,
that the presence of an object is made known to us; so
upon paper it is necessary, in order that the delineation
may appear real, to represent fully all the shades and shad-
ows that would be seen upon the object itself. In this sec-
tion I propose to illustrate, by a few plain examples, the
simple elementary principles upon which shading, in archi-
tectural subjects, is based. The necessary knowledge of
drawing, preliminary to this subject, is treated of in Section
XV., from Arts. 487 to 498.
563. — The Inclination of the JLlne of Shadow. — This
is always, in architectural drawing, 45 degrees, both on the
elevation and on the plan ; and the sun is supposed to be
behind the spectator, and over his left shoulder. This can
be illustrated by reference to Fig. 420, in which A repre-
sents a horizontal plane, and B and C two vertical planes
placed at right angles to each other. A represents the plan,
C the elevation, and B a vertical projection from the eleva-
tion. In finding the shadow of the plane B, the line a b is
drawn at an angle of 45 degrees with the horizon, and the
liner^ at the same angle with the vertical planed. The
plane B being a rectangle, this makes the true direction of
the sun's rays to be in a course parallel to db, which direc-
tion has been proved to be at an angle of 35 degrees and
16 minutes with the horizon. It is convenient, in shading,
to have a set-square with the two sides that contain the
CONVENTIONAL PLANES OF SHADOW.
597
right angle of equal length; this will make the two acute
angles each 45 degrees, and will give the requisite bevel
when worked upon the edge of the T-square. One reason
why this angle is chosen in preference to another is that
when shadows are properly made upon the drawing by it,
the depth of every recess is more readily known, since the
breadth of shadow and the depth of the recess will be equal.
To distinguish between the terms shade and shadow, it will
be understood that all such parts of a body as are not e5c-
posed to the direct action of the sun's rays are in shade ;
while those parts which are deprived of light by the inter-
position of other bodies are in shadow.
564. — To Find the Line of Shadow 011 Mouldings and
other Horizontally Straight Projections. — Figs. 421, 422,
423, and 424 represent various mouldings in elevation, re-
turned at the left, in the usual manner of mitering around a
projection. A mere inspection of the figures is sufficient to
see how the line of shadow is obtained, bearing in mind that
the ray a b is drawn from the projections at an angle of 45
598
SHADOWS.
degrees. When there is no return at the end, it is neces-
sary to draw a section, at any place in the length of the
mouldings, and find the line of shadow from that.
565. — To Find tlie Line of Shadow Cast by a Shelf. — In
Fig. 425, A is the plan and B is the elevation of a shelf
attached to a wall. From a and c draw a b and c d, accord-
ing to the angle previously directed ; from b erect a per-
pendicular intersecting c d <& d\ from d draw de parallel to
FIG. 421,
FIG. 422.
FIG. 423.
FIG. 424.
the shelf; then the lines cd and de will define the shadow
cast by the shelf. There is another method of finding the
shadow, without the plan A. Extend the lower line of the
shelf to //and make cf equal to the projection of the shelf
from the wall ; from/ draw fg at the customary angle, and
from c drop the vertical line eg intersecting fg^g\ from
g draw ge parallel to the shelf, and from c draw c d at the
usual angle; then the lines cd and de will determine the
extent of the shadow as before.
SHADOWS OF STRAIGHT AND OBLIQUE SHELVES. 599
.566.— To Find the Shadow Cast by a Shelf which i*
Wider at one End than at the Other. — In Fig. 426, A is the
plan, and B the elevation. Find the point d, as in the pre-
B
FIG. 425.
vious example, and from any other point in the front of the
shelf, as a, erect the perpendicular a e ; from a and e draw a b
and e c, at the proper angle, and from b erect the perpendicu-
FIG. 426.
lar be, intersecting ec in c\ from d, through c, dra\v do\
then the lines id and do will give the limit of the shadow
cast by the shelf.
6oo
SHADOWS.
567.— To Find tlie Shadow of a Shelf having one End
Aeute or Obtuse Angled. — Fig. 427 shows the plan and ele-
vation of an acute-angled shelf. Find the line eg as before ;
FIG. 427.
from a erect the perpendicular ab\ join b and e\ then be
and eg will define the boundary of shadow.
568.— To Find the shadow Cast by an Inclined Shelf.—
In Fig. 428 the plan and elevation of such a shelf are shown,
having also one end wider than the other. Proceed as di-
FIG 428.
rected for finding the shadows of Fig. 426, and find the points
^/and c ; then ad and dc will be the shadow required. If
the shelf had been parallel in width on the plan, then the
line dc would have been parallel with the shelf a b.
SHADOWS OF INCLINED AND CURVED SHELVES. 6oi
569.— To Find the Shadow Cast by a Shelf Inclined in
its Vertical Section either Upward or Downward. — From
a (Figs. 429 and 430) draw a b at the usual angle, and from b
draw be parallel with the shelf; obtain the point e by draw-
FIG. 429.
FIG. 430.
ing -a line from d at the usual angle. In Fig. 429 join e and
i ;. then ic and cc will define the shadow. In Fig. 430, from
o draw oi parallel with the shelf ; join i and e ; then ie and
cc will be the shadow required.
The projections in these several examples are bounded
FIG. 431-
FIG. 432.
by straight lines ; but the shadows of curved lines may be
found in the same manner, by projecting shadows from sev-
eral points in the curved line, and tracing the curve .of
shadow through these points. (Figs. 431 and 432.)
602
SHADOWS.
570.— To Find the Shallow of a Shelf having its Front
Edge, or End, Curved on the Plan, — In Figs. 431 and 432
A and A show an example of each kind. From several
points, as a, a, in the plan, and from the corresponding- points
0, o in the elevation, draw rays and perpendiculars intersect-
FIG. 433.
ing at e, <:, etc. ; through these points of intersection trace
the curve, and it will define the shadow.
57 L— To Find the Shadow of a Shelf Curved in the Ele-
vation.— In Fig. 433 find the points of intersection, e, c and
FIG. 434
e, as in the last examples, and a curve traced through them
will define the shadow.
The preceding examples show how to find shadows when
cast upon a vertical plane ; shadows thrown upon curved sur-
• faces are ascertained in a similar manner. (Fig. 434.)
SHADOW UPON AN INCLINED WALL.
603
572.— To Find the Shadow Cast upon a Cylindrical
Wall by a Projection of any Kind. — By an inspection of
Fig. 434, it will be seen that the only difference between this
and the last examples is that the rays in the plan die against
the circle ab, instead of a straight line.
573.— To Find the Shadow Ca§t by a Shelf upon an In-
clined Wall. — Cast the ray ab (Fig. 435) from the end of the
shelf to the face of the wall, and from b draw be parallel to
the shelf; cast the ray de from the end of the shelf; then
the lines de and ec will define the shadow.
FIG. 436.
These examples might be multiplied, but enough has
been given to illustrate the general principle by which shad-
ows in all instances are found. Let us attend now to the
application of this principle to such familiar objects as are
likely to occur in practice.
004 SHADOWS.
574.— To Find the Shadow of a Projecting Horizontal
Ream. — From the points a, a, etc. (Fig. 436), cast rays upon
the wall ; the intersections e, e, e of those rays with the per-
pendiculars drawn from the plan will define the shadow. If
the beam be inclined, either on the plan or elevation, at any
angle other than a right angle, the difference in the manner
FIG. 437.
of proceeding can be seen by reference to the preceding
examples of inclined shelves, etc.
575. — To Find the Shadow in a Recess. — From the point
a (Fig. 437) in the plan, and b in the elevation, draw the rays
acand be; from c erect the perpendicular ce, and from e
FIG. 438.
draw the horizontal line ed; then the lines r^.and ed will
show the extent of the shadow. This applies only where
the back of the recess is parallel with the face of the wall.
576.— To Find the Shadow in a Rece§§, when the Face
of the Wall Is Inclined, and the Baek of the Recess i§
Vertical. — In Fig. 438, A shows the section and B the eleva-
SHADOW IN A FIREPLACE.
605
tion of a recess of this kind. From b, and from any other
point in the line ba, as a, draw the rays be and ae ; from c,
a, and e draw the horizontal lines eg, af, and eh; from d
FIG. 439.
and /cast the rays di and ///; from i, through h, draw is;
then s i and ig will define the shadow.
577. TO Find the Shadow in a Fireplace. — From a and
b (Fig. 439) cast the rays a c and b e, and from c erect the
FIG. 440.
perpendicular cc\ from c draw the horizontal line eo, and
join o and </; then c c, eo, and <?</ will give the extent of the
shadow.
6o6
SHADOWS.
578.— To Find I lie Shadow of a moulded Window-Lin-
tel.— Cast rays from the projections a, o, etc., in the plan
(Fig. 440), and d, e, etc., in the elevation, and draw the usual
perpendiculars intersecting the rays at z, i, and i ; these in-
tersections connected, and horizontal lines drawn from them,
will define the shadow. The shadow on the face of the lin-
tel is found by casting a ray back from i to s, and drawing
the horizontal line s n.
579. — To Find the Shadow Cast by the No§ing of a Step.
— From a (Fig. 441) and its corresponding point c, cast the
FIG. 441.
rays a b and cd, and from b erect the perpendicular b d; tan-
gical to the curve at e cast the ray ef, and from c drop the
perpendicular e o, meeting the mitre-line ag in o ; cast a ray
from o to /', and from /erect the perpendicular if\ from /t
draw the ray // k ; from /to d and from d to k trace the
curve as shown in the figure ; from k and h draw the hori-
zontal lines kn and hs\ then the limit of the shadow will be
completed.
580.— To Find the Shadow Thrown by a Pede§tal upon
Step§. — From a (Fig. 442) in the plan, and from c in the ele-
vation, draw the rays ab and c e ; then ao will show the ex-
SHADOWS ON STEPS AND COLUMNS.
607
tent of the shadow on the first riser, as at A ; fg will deter-
mine the shadow on the second riser, as at B\ cd gives the
amount of shadow on the first tread, as at £7, and // i that on
the second tread, as at D ; which completes the shadow of
FIG. 442.
the left-hand pedestal, both on the plan and elevation. A
mere inspection of the figure will be sufficient to show how
the shadow of the right-hand pedestal is obtained.
FIG. 443.
FIG. 444.
681.— To Find the Shadow Thrown on a Column by a
Square Abacus. — From a and b (Fig. 443) draw the rays ac
and b e, and from c erect the perpendicular c e ; tangical to
the curve at d draw the ray df, and from //, corresponding
to /in the plan, draw the ray ho; take any point between a
and fy as i, and from this, as also from a corresponding point
6o8
SHADOWS.
n, draw the rays ir and ns ; from r and from d erect the
perpendiculars rs and do-, through the points e, s, and o
trace the curve as shown in the figure ; then the extent of
the shadow will be defined.
m
X.
N/^
A. /"
J
^^^^^ ^TlWk
FIG. 445.
582. — To Find the Shadow Tin-own on a Column by a
Circular Abacus. — This is so nearly like the last example
that no explanation will be necessary, farther than a refer-
ence to the preceding article.
SHADOWS ON THE CAPITAL OF A COLUMN".
609
583.— To Find the Shadows on the Capital of a Columii.
— This may be done according- to the principles explained
in the examples already given ; a quicker way of doing it,
however, is as follows : if we take into consideration one
ray of light in connection with all those perpendicularly
under and over it, it is evident that these several rays would
form a vertical plane, standing at an angle of 45 degrees
with the face of the elevation. Now we may suppose the
column to be sliced, so to speak, with planes of this nature — •
cutting it in the lines a b, c d, etc. (Fig. 445), and, in the ele-
FIG. 446.
vation, find by squaring up from the plan, the lines of section
which these planes would make thereupon. For instance :
in finding upon the elevation the line-of section a I), the plane
cuts the ovolo at e, and therefore / will be the correspond-
ing point upon the elevation ; h corresponds with g, i withy,
o with s, and / with b. Now, to find the shadows upon tfcis
line of section, cast from m the ray m «, from // the ray h o,
etc. ; then that part of the section indicated by the letters
m f i n, and that part also between // and o will be under
6io
SHADOWS.
shadow. By an inspection of the figure, it will be seen that
the same process is applied to each line of section, and in
that way the points /, r, t, u, v, ?v, x, as also i, 2, 3, etc., are
FIG. 447.
successively found, and the lines of shadow traced through
them.
Fig. 446 is an example of the same capital with all the
shadows finished in accordance with the lines obtained on
Fig. 445.
SHADOW OF A COLUMN ON A WALL.
584.— To Find the Shadow Thrown on a Vertical Wall
l>y a Column and Entablature Standing in Advance of §aid
Wall. — Cast rays from a and b (Fig. 447), and find the point
c as in the previous examples ; from d draw the ray de, and
from e the horizontal line ef\ tangical to the curve at g and
h draw the rays gj and h i, and from i and j erect the per-
pendiculars il and/£; from m and n draw the rays mf and
nk, and trace the curve between £and/; cast a ray from o to
/, a vertical line from/ to s, and through s draw the horizon-
tal line s t ; the shadow as required will then be completed.
FIG. 448.
Fig. >| /| K is an example of the same kind as the last, with
all the shadows filled in, according to the lines obtained in
the preceding figure.
585. — Shadows on a Cornice. — Figs. 449 and 450 are
examples of the Tuscan cornice. The manner of obtaining
the shadows is evident.
586. — Reflected Light. — In shading, the finish and life of
an object depend much on reflected light. This is seen to
advantage in Fig. 446, and on the column in Fig. 448. Re-
6l2
SHADOWS.
fleeted rays are thrown in a direction exactly the reverse
of direct rays ; therefore, on that part of an object which is
subject to reflected light, the shadows are reversed. The
FIG. 449.
fillet of the ovolo in Fig. 446 is an example of this. On the
right hand side of the column, the^ace of the fillet is much
darker than the cove directly under it. The reason of this
FIG. 450,
is, the face of the fillet is deprived both of direct and re-
flected light, whereas the cove is subject to the latter. Other
instances of the effect of reflected light will be seen in the
other examples.
CONTENTS.
PART I.
SECTION I. — ARC H ITECTU RE.
Art. 1. Building defined, p. 5.— 2. Antique Builcings ; Tower of Babel,
p. 5. — 3. Ancient Cities and Monuments, p. 6. — 4. Architecture in Greece,
p. 6. — 5. Architecture in Rome, p. 7. — 6. Rome and Greece, p. 8. — 7. Ar-
chitecture debased, p. 9. — §. The Ostrogoths, p. 9. — 9. The Lombards, p. 10.
— 1O. The Byzantine Architects, p. 10. — 11. The Moors, p. 10. — 12. The
Architecture of England, p. n. — 13. Architecture Progressive, p. 12. — 14.
Architecture in Italy, p. 12. — 15. The Renaissance, p. 13. — 16. Styles of Ar-
chitecture, p. 13. — 17. Orders, p. 14.— 18. The Stylobatc, p. 14. — 19. The
Column, p. 14.— 2O. The Entablature, p. 14.— 21. The Base, p. 14—22.
The Shaft, p. 15.— 23. The Capital, p. 15.— 24. The Architrave, p. 15.— 25.
The Frieze, p. 15.— 26. The Cornice, p. 15.— 27. The Pediment, p. 15.— 28.
The Tympanum, p. 15. — 29. The Attic, p. 15. — 3O. Proportions in an Order,
p. 15.— 31. Grecian Styles, p. 16.— 32. The Doric Order, p. 16.— 33. The
Intercolumniation, p. 17.— 34. The Doric Order, p. 19 — 35. The Ionic
Order, p. 19. — 36. The Intercolumniation, p. 20. — 37. To Describe the Ionic
Volute, p. 20. — 38. The Corinthian Order, p. 23. — 39. Persians and Carya-
tides, p. 24. — 4O. Persians, p. 24. — 41, Caryatides, p. 26. — 42. Roman
Styles, p. 26. — 43. Grecian Orders modified by the Romans, p. 27. — 44. The
Tuscan Order, p. 30.— 45. Egyptian Style, p. 30. — 46. Building in General,
p. 33.— 47. Expression, p. 35. — 48. Durability, p. 37.— 49. Dwelling-
Houses, p. 37. — 5O. Arranging the Stairs and Windows, p. 42. — 51. Prin-
ciples of Architecture, p. 44.— 52. Arrangement, p. 44. — 53. Ventilation, p.
45. — 54. Stability, p. 45. — 55. Decoration, p. 46.— 56. Elementary Parts of
a Building, p. 46. — 57. The Foundation, p. 47. — 58. The Column, or Pillar,
p. 47.— 59. The Wall, p. 48. — 6O. The Reticulated Walls, p. 49.— 61. The
Lintel, or Beam, p. 49. — 62. The Arch, p. 50. — 63. Hookc's Theory of an
Arch, p. 50. — 64. Gothic Arches, p. 51. — 65. Arch : Definitions ; Principles,
p. 52.— 66. An Arcade, p. 52.— 67. The Vault, p. 52.— 68. The Dome, p. 53.
—69. The Roof, p. 54.
614 CONTENTS.
SECTION II.— CONSTRUCTION.
Art. TO. Construction Essential, p. 56. — 71. Laws of Pressure, p. 57. —
72. Parallelogram of Forces, p. 59.— 73. The Resolution of Forces, p. 59. —
74. Inclination of Supports Unequal, p. 60. — 75. The Strains Exceed the
Weights, p. 61. — 76. Minimum Thrust of Rafters, p. 62. — 77. Practical
Method of Determining Strains, p. 62. — 78. Horizontal Thrust, p. 63. — 79.
Position of Supports, p. 65. — 8O. The Composition of Forces, p. 66. — 81.
Another Example, p. 67. — 82. Ties and Struts, p. 68. — 83. To Distinguish
Ties from Struts, p. 69. — 84. Another Example, p. 70. — 85. Centre of Gravity,
p. 7I._86. Effect of the Weight of Inclined Beams, p. 72. — 87. Effect of
Load on Beam, p. 74. — 88. Effect on Bearings, p. 75. — 89. Weight-Strength,
p. 76. — 9O. Quality of Materials, p. 76. — 91. Manner of Resisting, p. 77. —
92. Strength and Stiffness, p. 78. — 93. Experiments : Constants, p. 78. — 94.
Resistance to Compression, p. 79. — 95. Resistance to Tension, p. 81. — 96.
Resistance to Transverse Strains, p. 83. — 97. Resistance to Compression, p.
85. — 98. Compression Transversely to the Fibres, p. 86. — 99. The Limit of
Weight, p. 86.— 1OO. Area of Post, p. 86.— 1O1. Rupture by Sliding, p. 87.
— 1O2. The Limit of Weight, p. 87.— 1O3. Area of Surface, p. 88.— 1O4.
Tenons and Splices, p. 88. — HO5. Stout Posts, p. 89.— IO6. The Limit of
Weight, p. 89. — 1®7. Area of Post, p. 90. — 1O8. Area of Round Posts, p.
90.— 1O9. Slender Posts, p. 91.— 11O. The Limit of Weight, p. 91.— 111.
Diameter of the Post; when Round, p. 92. — 112. Side of Post: when
Square, p. 93.— 113. Thickness of a Rectangular Post, p. 95. — 114. Breadth
of a Rectangular Post, p. 95.— 115. Resistance to Tension, p. 96. — 116.
.The Limit of Weight, p. 96.— 117. Sectional Area, p. 97.— 118. Weight of
the Suspending Piece Included, p. 98. — 119. Area of Suspending Piece,
p. 99.
RESISTANCE TO TRANSVERSE STRAINS.
Art. H2O. Transverse Strains: Rupture, p. 99. — H21L. Location of Mor-
tises, p. 100. — 122. Transverse Strains : Relation of Weight to Dimensions,
p. loi— 123. Safe Weight : Load at Middle, p. 103.— 124. Breadth of Beam
with Safe Load, p. 104. — fl25. Depth of Beam with Safe Load, p. 104. — 126.
Safe Load at any Point, p. 105. — 127. Breadth or Depth : Load at any Point,
p. 106. — 128. Weight Uniformly Distributed, p. 107. — 129. Breadth or
Depth: Load Uniformly Distributed, p. 108. — 13O. Load per Foot Super-
ficial, p. 109. — 131. Levers : Load at one End, p. no. — 132. Levers : Breadth
or Depth, p. in. — 133. Deflection: Relation to Weight, p. 112. — 134. De-
flection: Relation t<) Dimensions, p. 112. — 135. Deflection : Weight when at
Middle, p. 114. — 136. Deflection : Breadth or Depth, Weight at Middle, p.
114.— 137. Deflection : When Weight is at Middle, p. 116.— 138. Deflection :
Load Uniformly Distributed, p. 116. — 139. Deflection : Weight when Uni-
formly Distributed, p. 117.— I4O. Deflection: Breadth or Depth, Load Uni-
formly Distributed, p. 117. — 14i. Deflection: When Weight is Uniformly
Distributed, p. 118. — 142. Deflection of Lever, p. 119. — 143. Deflection of
a Lever: Load at End, p. 120. — 144. Deflection of a Lever : Weight when at
End, p. 120. — 145. Deflection of a Lever : Breadth or Depth, Load at End,
CONTENTS. 615
p. i2i.— 146. Deflection of Levers: Weight Uniformly Distributed, p. 121.—
147. Deflection of Levers with Uniformly Distributed Load, p. 122.. — 148.
Deflection of Levers: Weight when Uniformly Distributed, p. 122. — 149.
Deflection of Levers : Breadth or Depth, Load Uniformly Distributed, p. 122.
CONSTRUCTION IN GENERAL.
Art. 15O. Construction : Object Clearly Denned, p. 123. — 151. Floors
Described, p. 124. — 152. Floor-Beams, p. 125. — 153. Floor-Beams for Dwell-
ings, p. 127. — 154. Floor-Beams for First-Class Stores, p. 128. — 155. Floor-
Beams : Distance from Centres, p. 129. — 156. Framed Openings for Chimneys
and Stairs, p. 130. — 157. Breadth of Headers, p. 130. — 15§. Breadth of
Carriage-Beams, p. 132. — 159. Breadth of Carriage-Beams Carrying Two
Sets of Tail-Beams, p. 134. — 16O. Breadth of Carriage-Beam with Well-Hole
at Middle, p. 136. — 161. Cross- Bridging, or Herring-Bone Bridging, p. 137.
—162. Bridging : Value to Resist Concentrated Loads, p. 137. — 163. Gird-
ers, p. 140. — 164. Girders : Dimensions, p. 141.
FIRE-PROOF TIMBER FLOORS.
Art. 165. Solid Timber Floors, p. 143. — 166. Solid Timber Floors for
Dwellings and Assembly-Rooms, p. 143. — 167. Solid Timber Floors for First-
Class Stores, p. 144. — 168. Rolled-Iron Beams, p. 145. — 169. Rolled-Iron
Beams: Dimensions; Weight at Middle, p. 146. — 17O. Rolled-Iron Beams:
Deflection when Weight is at Middle, p. 147. — 171. Rolled-Iron Beams:
Weight when at Middle, p. 148. — 172. Rolled-Iron Beams: Weight at any
Point, p. 148. — 173. Rolled-Iron Beams: Dimensions ; Weight at any Point,
p ^g. — B.74. Rolled-Iron Beams : Dimensions ; Weight Uniformly Distrib-
uted, p. 149. — 175. Rolled-Iron Beams : Deflection ; Weight Uniformly Dis-
tributed, p. 150. — 176. Rolled-Iron Beams : Weight when Uniformly Distrib-
uted, p. 151. — 177. Rolled-Iron Beams: Floors of Dwellings or Assembly-
Rooms, p. 151. — 178. Rolled-Iron Beams : Floors of First-Class Stores, p.
152. — 179. Floor-Arches: General Considerations, p. 153. — 18O. Floor-
Arches: Tie-Rods; Dwellings, p. 153. — 181. Floor-Arches: Tie-Rods;
First Class Stores, p. 153.
TUBULAR IRON GIRDERS.
Art. 182. Tubular Iron Girders: Description, p. 154.— 183. Tubular
Iron Girders : Area of Flanges ; Load at Middle, p. 154. — 184. Tubular Iron
Girders : Area of Flanges ; Load at any Point, p. 155. — 185. Tubular Iron
Girders : Area of Flanges ; Load Uniformly Distributed, p. 156. — 186. Tu-
bular Iron Girders: Shearing Strain, p. 157. — 187. Tubular Iron Girders:
Thickness of Web, p. 158. — 188. Tubular Iron Girders for Floors of Dwell-
ings, Assembly-Rooms, and Office Buildings, p. 159.— 189. Tubular Iron
Girders for Floors of First-Class Stores, p. 160.
CAST-IRON GIRDERS.
Art. 19O. Cast-Iron Girders: Inferior, p. 161. — 191. Cast-Iron Girder:
Load at Middle, p. 161. — 192. Cast-Iron Girder: Load Uniformly Distributed,
6l6 CONTENTS.
p. 163. — 193. Cast-Iron Bowstring Girder, p. 163. — 194. Substitute for the
Bowstring Girder, p. 163.
FRAMED GIRDERS.
Art. 195. Graphic Representation of Strains, p. 165. — 196. Framed
Girders, p. 166. — 197. Framed Girder and Diagram of Forces, p. 167. — 198.
Framed Girders : Load on Both Chords, p. 171. — 199. Framed Girders: Di-
mensions of Parts, p. 173.
PARTITIONS.
Art. 2OO. Partitions, p. 174. — 2O1. Examples of Partitions, p. 175.
ROOFS.
Art. 2O2. Roofs, p. 178.— 2O3. Comparison of Roof-Trusses, p. 178.—
2O4. Force Diagram : Load upon Each Support, p. 179. — 2O5. Force Dia-
gram for Truss in Fig. 59, p. 179. — 2O6. Force Diagram for Truss in Fig. 60,
p. 180. — 2O7. Force Diagram for Truss in JFtg.Gi, p. 181. — 2O8. Force Dia-
gram for Truss in Fig. 63, p. 183. — 299. Force Diagram for Truss in Fig. 64,
p. 184.— 21O. Forc« Diagram for Truss in Fig. 65, p. 185.— 211. Force Dia-
gram for Truss in Fig. 66, p. 186.— 212. Roof-Truss : Effect of Elevating the
Tie-Beam, p. 187. — 213. Planning a Roof, p. 188. — 214. Load upon Roof-
Truss, p. 189. — 215. Load on Roof per Superficial Foot, p. 189.— 216. Load
upon Tie-Beam, p. 190.— 217. Roof Weights in Detail, p. 191.— 21§. Load
per Foot Horizontal, p. 192. — 219. Weight of Truss, p. 192. — 22O. Weight
of Snow on Roofs, p. 193. — 221. Effect of Wind on Roofs, p. 193.— 222.
Total Load per Foot Horizontal, p. 197. — 223. Strains in Roof Timbers
Computed, p. 198. — 224. Strains in Roof Timbers Shown Geometrically, p.
199. — 225. Application of the Geometrical System of Strains, p. 202. — 226.
Roof Timbers : the Tie-Beam, p. 204.— 227. The Rafter, p. 205.— 228. The
Braces, p. 208. — 229. The Suspension Rod, p. 210. — 23O. Roof-Beams,
Jack-Rafters, and Purlins, p. 211. — 231. Five Examples of Roofs, p. 212. —
232. Roof-Truss with Elevated Tie-Beam, p. 214. — 233. Hip-Roofs : Lines
and Bevels, p. 215. — 234. The Backing of the Hip-Rafter, p. 216.
DOMES.
Art. 235. Domes, p. 216.— 236. Ribbed Dome, p. 217.— 237. Domes:
Curve of Equilibrium, p. 218. — 238. Cubic Parabola Computed, p. 219. —
239. Small Domes over Stairways, p. 220. — 24O. Covering for a Spherical
Dome, p. 221. — 241. Polygonal Dome : Form of Angle-Rib, p. 223.
BRIDGES.
Art. 242. Bridges, p. 223.— 243. Bridges : Built-Rib, p. 224.— 244.
Bridges : Framed Rib, p. 226. — 245. Bridges : Roadway, p. 227. — 246.
Bridges : Abutments, p. 227. — 247. Centres for Stone Bridges, p. 229. — 248.
Arch Stones : Joints, p. 223.
JOINTS.
Art. 249. Timber Joints, p. 234.
CONTENTS. 6i;
SECTION III.— STAIRS.
Art. 250. Stairs : General Requirements, p. 240. — 251. The Grade of
Stairs, p. 241.— 252. Pitch-Board : Relation of Rise to Tread, p. 242. — 253.
Dimensions of the Pitch-Board, p. 247. — 254. The String of a Stairs, p. 247.
— 255. Step and Riser Connection, p. 248.
PLATFORM STAIRS
Art. 256. Platform Stairs : the Cylinder, p. 248.— 257. Form of Lower
Edge of Cylinder, p. 249. — 25§. Position of the Balusters, p. 250. — 259. Wind-
ing Stairs, p. 251. — 260. Regular Winding Stairs, p. 251.— 261. Winding
Stairs : Shape and Position of Timbers, p. 252. — 262. Winding Stairs with
Flyers: Grade of Front String, p. 253.
HAND-RAILING.
Art. 263. Hand-Railing for Stairs, p. 256.— 264. Hand-Railing : Defini-
tions ; Planes and Solids, p. 257. — 265. Hand-Railing: Preliminary Consider-
ations, p. 258. — 266. A Prism Cut by an Oblique Plane, p. 259. — 267. Form
of Top of Prism, p. 259.— 26§. Face-Mould for Hand-Railing of Platform
Stairs, p. 264. — 269. More Simple Method for Hand-Rail to Platform Stairs,
p. 267. — 27O. Hand-Railing for a Larger Cylinder, p. 271. — 271. Face-
Mould without Canting the Plank, p. 272. — 272. Railing for Platform Stairs
where the Rake meets the Level, p. 272. — 273. Application of Face-Moulds
to Plank, p. 273. — 274. Face-Moulds for Moulded Rails upon Platform
Stairs, p. 274. — 275. Application of Face-Moulds to Plank, p. 275. — 276.
Hand-Railing for Circular Stairs, p. 278. — 277. Face-Moulds for Circular
Stairs, p. 282.— 27§. Face-Moulds, for Circular Stairs, p. 285—279. Face-
Moulds for Circular Stairs, again, p. 287. — 280. Hand-Railing for Winding
Stairs, p. 289.— 2§1. Face-Moulds for Windjng Stairs, p. 290.— 2§2. Face-
Moulds for Winding Stairs, again, p. 293. — 283. Face-Moulds : Test of Accu-
racy, p. 295.— 284. Application of the Face-Mould, p. 297. — 285. Face-Mould
Curves are Elliptical, p. 301. — 286. Face-Moulds for Round Rails, p. 303. —
287. Position of the Butt Joint, p. 303.— 288. Scrolls for Hand-Rails: Gen-
eral Rule for Size and Position of the Regulating Square, p. 308. — 289. Cen-
tres in Regulating Square, p. 308. — 29O. Scroll for Hand-Rail Over Curtail
Step, p. 309.— 291. Scroll for Curtail Step, p. 310.— 292. Position of Balus-
ters Under Scroll, p. 310— 293. Falling-Mould for Raking Part of Scroll, p.
310. — 294. Face-Mould for the Scroll, p. 311. — 295. Form of Newel-Cap
from a Section of the Rail, p. 312.— 296. Boring for Balusters in a Round Rail
before it is Rounded, p. 313.
SPLAYED WORK.
Art. 297. The Bevels in Splayed Work, p. 314.
SECTION IV. — DOORS AND WINDOWS.
DOORS.
Art. 298. General Requirements, p. 315. — 299. The Proportion between
Width and Height, p. 315.— 3OO. Panels, p. 316.— 3O1. Trimmings, p. 317.
— 302. Hanging Doors, p. 317.
6l8 CONTENTS.
WINDOWS.
Art. 3O3. Requirements for Light, p. 317. — 3O4. Window Frames, p.
318.— 3O5. Inside Shutters, p. 319.— 3O6. Proportion: Width and Height,
p. 319.— 3O7. Circular Heads, p. 320.— 3O8. Form of Soffit for Circular Win-
dow Heads, p. 321.
SECTION V.— MOULDINGS AND CORNICES.
MOULDINGS.
Art. 309. Mouldings, p. 323. — 31 0. Characteristics of Mouldings, p.
324. — 311. A Profile, p. 326. — 312. The Grecian Torus and Scotia, p. 326. —
313. The Grecian Echinus, p. 327. — 314. The Grecian Cavetto, p. 327. —
315. The Grecian Cyma- Recta, p. 327. — 316. The Grecian Cyma-Reversa,
p. 328.— 3S.7. Roman Mouldings, p. 329. — 318. Modern Mouldings, p. 331.
CORNICES.
Art. 3H9. Designs for Cornices, p. 335. — 32O. Eave Cornices Propor-
tioned to Height of Building, p. 335. — 321. Cornice Proportioned to a given
Cornice, p. 342. — 322. Angle Bracket in a Built Cornice, p. 343. — 323. Rak-
ing Mouldings Matched with Level Returns, p. 344.
PART II.
SECTION VI.-GEOMETRY.
Art. 324. Mathematics Essential, p. 347. — 325. Elementary Geometry,
p. 347. — 326. Definition— Right Angles, p. 348. — 327. Definition — Degrees
in a Circle, p. 348. — 328. Definition — Measure of an Angle, p. 348.— 329.
Corollary — Degrees in a Right Angle, p. 348.— 33O. Definition — Equal
Angles, p. 349. — 331. Axiom — Equal Angles, p. 349. — 332. Definition —
Obtuse and Acute Angles, p. 349. — 333. Axiom — Right Angles, p. 349. —
334. Corollary — Two Right Angles, p. 349. — 335. Corollary — Four Right
Angles, p. 349. — 336. Proposition — Equal Angles, p. 350.— 337. Propo-
sition— Equal Triangles, p. 350. — 338. Proposition — Angles in Isosceles
Triangle, p. 351. — 339. Proposition — Diagonal of Parallelogram, p. 351- —
34O. Proposition — Equal Parallelograms, p. 352. — 341. Proposition — Paral-
lelograms Standing on the Same Base, p. 352. — 342. Corollary — Parallelo-
gram and Triangle, p. 353. — 343. Proposition — Triangle Equal to Quadrangle,
p. 353. — 344. Proposition — Opposite Angles Equal, p. 354. — 345. Proposi-
tion— Three Angles of Triangle Equal to Two Right Angles, p. 354. — 346.
Corollary— Right Angle in Triangle, p. 354.— 347. Corollary— Half a Right
CONTENTS. 619
Angle, p. 355.— 348. Corollary— Right Angle in a Triangle, p. 355.— 349.
Corollary — Two Angles Equal to Right Angle, p. 355.— 35O. Corollary — Two
Thirds of a Right Angle, p. 355. — 351. Corollary — Equilateral Triangle, p. 355.
— 352. Proposition — Right Angle in Semi-circle, p. 355. — 353. Proposition—-
The Square of the Hypothenuse Equal to the Squares of the Sides, p. 355. —
354. Proposition — Equilateral Octagon, p. 357. — 355. Proposition — Angle
at the Circumference of a Circle, p. 358.— 356. Proposition — Equal Chords
give Equal Angles, p. 358. — 357. Corollary of Equal Chords, p. 359.— 35§.
Proposition— Angle Formed by a Chord and Tangent, p. 359. — 359. Propo-
sition— Areas of Parallelograms, p. 360. — 360. Proposition — Triangles ot
Equal Altitude, p. 361. — 361. Proposition— Homologous Triangles, p. 362. —
362. Proposition — Parallelograms of Chords, p. 363. — 363. Proposition — •
Sides of Quadrangle, p. 364.
SECTION VII.— RATIO, OR PROPORTION.
Art. 364. Merchandise, p. 366.— 365. The Rule of Three, p. 366.—
366. Couples: Antecedent, Consequent, p. 367. — 367. Equal Couples : an
Equation, p. 367. — 36§. Equality of Ratios, p. 367. — 369. Equals Multiplied
by Equals Give Equals, p. 367. — 37O. Multiplying an Equation, p. 368. — 371.
Multiplying and Dividing one Member of an Equation : Cancelling, p. 368. —
372. Transferring a Factor, p. 369. — 373. Equality of Product : Means and
Extremes, p. 369. — 374. Homologous Triangles Proportionate, p. 370. —
375. The Steelyard, p. 371.— 376. The Lever Exemplified by the Steelyard,
p. 372. — 377. The Lever Principle Demonstrated, p. 375. — 378. Any One or
Four Proportionals may be Found, p. 377.
SECTION VIII.— FRACTIONS.
Art. 379. A Fraction Defined, p. 378.— 38O. Graphical Representation
of Fractions : Effect of Multiplication, p. 378. — 381. Form of Fraction
Changed by Division, p. 380. — 382. Improper Fractions, p. 380. — 383. Re-
duction of Mixed Numbers to Fractions, p. 381. — 384. Division Indicated by
the Factors put as a Fraction, p. 381.— 385. Addition of Fractions having Like
Denominators, p. 382. — 386. Subtraction of Fractions of Like Denominators,
p> 383. — 387. Dissimilar Denominators Equalized, p. 383. — 388. Reduction
of Fractions to their Lowest Terms, p. 384. — 389. Least Common Denomina-
tor, p. 384. — 390. Least Common Denominator Again, p. 385. — 391. Frac-
tions Multiplied Graphically, p. 386.— 392. Fractions Multiplied Graphically
Again, p. 387.— 393. Rule for Multiplication of Fractions, and Example, p.
387.— 394. Fractions Divided Graphically, p. 388.— 395. Rule for Division
of Fractions, p. 389.
SECTION IX.— ALGEBRA.
Art. 396. Algebra Defined, p. 392.— 397. Example: Application, p.
393> — 398. Algebra Useful in Constructing Rules, p. 394. — 399. Algebraic
Rules are General, p. 394.— 4OO. Symbols Chosen at Pleasure, p. 395.— 4O1.
Arithmetical Processes Indicated by Signs, p. 396. — 4O2. Examples in Addi-
C2O CONTENTS.
tion and Subtraction : Cancelling, p. 398. — 403. Transferring a Symbol to the
Opposite Member, p. 399. — 4O4. Signs of Symbols to be Changed when they
are to be Subtracted, p. 400. — 4O5. Algebraic Fractions, Added and Sub-
tracted, p. 403.— 4O6. The Least Common Denominator., p. 404.— 4O7. Alge-
braic Fractions Subtracted, p. 405. — 4O§. Graphical Representation of Multi-
plication, p. 408. — 4O9. Graphical Multiplication : Three Factors, p. 408. —
41O. Graphic Representation : Two and Three Factors, p. 409. — 411. Graph-
ical Multiplication of a Binomial, p. 409. — 412. Graphical Squaring of a
Binomial, p. 410.— 413. Graphical Squaring of the Difference of Two Fac-
tors, p. 412. — 414. Graphical Product of the Sum and Difference of Two
Quantities, p. 413. — 415. Plus and Minus Signs in Multiplication, p. 415, —
416. Equality of Squares on Hypothenuse and Sides of Right-Angled Tri-
angle, p. 416. — 417*. Division the Reverse of Multiplication, p. 418. — 418.
Division: Statement of Quotient, p. 419. — 419. Division: Reduction, p. 419.
— 420. Proportionals : Analysis, p. 421. — 421. Raising a Quantity to any
Power, p. 423. — 422. Quantities with Negative Exponents, p. 423.— 423.
Addition and Subtraction of Exponential Quantities, p. 424. — 424. Multipli-
cation of Exponential Quantities, p. 424. — 425. Division of Exponential
Quantities, p. 424. — 426. Extraction of Radicals, p. 425. — 427. Logarithms,
p. 425. — 428. Completing the Square of a Binomial, p. 429.
PROGRESSION.
Art. 429. Arithmetical Progression, p. 432. — 43O. Geometrical ProgreS'
sion, p. 435.
SECTION X.--POLYGONS.
Art. 431. Relation of Sum and Difference of Two Lines, p. 439. — 432.
Perpendicular, in Triangle of Known Sides, p. 440. — 433. Trigon : Radius of
Circumscribed and Inscribed Circles : Area, p. 443. — 434. Tetragon : Radius
of Circumscribed and Inscribed Circles: Area, p. 446. — 435. Hexagon : Ra-
dius ot Circumscribed and Inscribed Circles : Area, p. 447. — 436. Octagon :
Radius of Circumscribed and Inscribed Circles : Area, p. 449. — 437. Dodec-
agon: Radius of Circumscribed and Inscribed Circles: Area, p. 452. — 438.
Hecadecagon : Radius of Circumscribed and Inscribed Circles : Area, p. 455.
— 439. Polygons : Radius of Circumscribed and Inscribed Circles : Area, p.
460. — 440. Polygons : Their Angles, p. 462. — 441. Pentagon: Radius of the
Circumscribed and Inscribed Circles: Area, p. 463. — 442. Polygons: Table
of Constant Multipliers, p. 465.
SECTION XL— THE CIRCLE.
Art. 443. Circles : Diameter and Perpendicular : Mean Proportional, p.
468. — 444. Circle : Radius from Given Chord and Versed Sine, p. 469. —
445. Circle : Segment from Ordinates, p. 470. — 446. Circle : Relation of
Diameter to Circumference, p. 472. — 447. Circle : Length of an Arc, p. 475.
— 448. Circle : Area, p. 475. — 449. Circle: Area of a Sector, p. 476. — 45O.
Circle : Area of a Segment, p. 477.
CONTENTS. 621
SECTION XII.— THE ELLIPSE.
Art. 451. Ellipse: Definitions, p. 481. — 452. Ellipse: Equations to the
Curve, p. 482. — 453. Ellipse : Relation of Axis to Abscissas of Axes, p. 484.
— 454. Ellipse : Relation of Parameter and Axes, p. 485. — 455. Ellipse :
Relation of Tangent to the Axes, p. 485. — 456. Ellipse: Relation of Tangent
with the Foci, p. 487. — 457. Ellipse : Relation of Axes to Conjugate Diam-
eters, p. 487.— 458. Ellipse : Area, p. 488.— 459. Ellipse : Practical Sugges-
tions, p. 489.
SECTION XIII.— THE PARABOLA.
Art. 46O. Parabola : Definitions, p. 492. — 461. Parabola : Equation to
the Curve, p. 493. — 462. Parabola : Tangent, p. 493. — 463. Parabola : Sub-
tangent, p. 496. — 464. Parabola : Normal and Subnormal, p. 496. — 465.
Parabola : Diameters, p. 497. — 466. Parabola : Elements, p. 499. — 467.
Parabola : Described Mechanically, p. 500. — 468. Parabola : Described from
Points, p. 502. — 469. Parabola : Described from Arcs, p. 503. — 47O. Para-
bola : Described from Ordinates, p. 504. — 471. Parabola : Described from
Diameters, p. 507. — 472. Parabola : Area, p. 509.
SECTION XIV.— TRIGONOMETRY.
Art. 473. Right-Angled Triangles: The Sides, p. 510.— 474. Right-
Angled Triangles : Trigonometrical Tables, p. 512. — 475* Right-Angled
Triangles: Trigonometrical Value of Sides, p. 516. — 476. Oblique- Angled
Triangles: Sines and Sides, p. 519. — 477. Oblique-Angled Triangles : First
Class, p. 520. — 478. Oblique-Angled Triangles: Second Class, p. 522. —
479. Oblique-Angled Triangles : Sum and Difference of Two Angles, p. 523.
— 48O. Oblique-Angled Triangles : Third Class, p. 526. — 481. Oblique-
Angled Triangles : Fourth Class, p. 528. — 482* Trigonometrical Formulae :
Right-Angled Triangles, p. 530. — 483. Trigonometrical Formula; : First
Class, Oblique, p. 531. — 484. Trigonometrical Formulae: Second Class,
Oblique, p. 532. — 485. Trigonometrical Formulae : Third Class, Oblique, p.
534.— 486. Trigonometrical Formulas : Fourth Class, Oblique, p. 534.
SECTION XV.— DRAWING
Art. 487. General Remarks, p. 536.— 488. Articles Required, p. 536.—
489. The Drawing-Board, p. 536.— 49O. Drawing- Paper, p. 537.— 491. To
Secure the Paper to the Board, p. 537.— 492. The T-Square, p. 539.— 493.
The Set-Square, p. 539. — 494. The Rulers, p. 540.— 495. The Instruments,
p. 540.— 496. The Scale of Equal Parts, p. 540.— 497. The Use of the Set-
Square, p.. 541. — 498. Directions for Drawing, p. 542.
622 CONTENTS.
SECTION XVI.— PRACTICAL GEOMETRY.
Art. 499. Definitions of Various Terms, p. 544.
PROBLEMS.
RIGHT LINES AND ANGLES.
Art. 5OO. To Bisect a Line, p. 549.— 5O1. To Erect a Perpendicular, p.
550.— 5O2. To let Fall a Perpendicular, p. 551.— 5O3. To Erect a Perpen-
dicular at the End of a Line, p. 551. — 5O4. To let Fall a Perpendicular near
the End of a Line, p. 553. — 5O5. To Make an Angle Equal to a Given Angle,
p. 553. — 5O6. To Bisect an Angle, p. 554. — 5O7. To Trisect a Right Angle,
p. 554. — 5O8. Through a Given Point to Draw a Line Parallel to a Given
Line, p. 555. — 509. To Divide a Given Line into any Number of Equal Parts,
P- 555-
THE CIRCLE.
Art. 5IO. To Find the Centre of a Circle, p. 556. — 511. At a Given
Point in a Circle to Draw a Tangent thereto, p. 557. — 512. The Same, with-
out making use of the Centre of the Circle, p. 557. — 513. A Circle and a
Tangent Given, to Find the Point of Contact, p. 558. — 514. Through any
Three Points not in a Straight Line to Draw a Circle, p. 559.— 515. Three
Points not in a Straight Line being Given, to Find a Fourth that Shall, with
the Three, Lie in the Circumference of a Circle, p. 559. — 516. To Describe
a Segment of a Circle by a Set-Triangle, p. 560. — 517. To Find the Radius of
an Ate of a Circle when the Chord and Versed Sine are Given, p. 561. — 518.
To Find the Versed Sine of an Arc of a Circle when the Radius and Chord
are Given, p. 561. — 519. To Describe the Segment of a Circle by Intersection
of Lines, p. 562. — 52O. Ordinates, p. 563. — 521. In a Given Angle to De-
scribe a Tanged Curve, p. 565. — 522. To Describe a Circle within any Given
Triangle, so that the Sides of the Triangle shall be Tangical, p. 566.— 523.
About a Given Circle to Describe an Equilateral Triangle, p. 566. — 524. To
Find a Right Line nearly Equal to the Circumference of a Circle, p. 566.
POLYGONS, ETC.
Art. 525. Upon a Given Line to Construct an Equilateral Triangle, p.
568.— 526. To Describe an Equi'lateral Rectangle, or Square, p. 568. — 527.
Within a Given Circle to Inscribe an Equilateral Triangle, Hexagon, or Dodec-
agon, p. 569. — 52§. Within a Square to Inscribe an Octagon, p. 570. — 529.
To Find the Side of a Buttressed Octagon, p. 571. — 5ilO. Within a Given
Circle to Inscribe any Regular Polygon, p. 572. — 531. Upon a Given Line to
Describe any Regular Polygon, p. 573. — 532. To Construct a Triangle whose
Sides shall be severally Equal to Three Given Lines, p. 575. — 533. To Con-
struct a Figure Equal to a Given Right-lined Figure, p. 575. — 534. To Make
a Parallelogram Equal to a Given Triangle, p. 576. — 535. A Parallelogram
being Given, to Construct Another Equal to it, and Having a Side Ecjual to a
CONTENTS. 623
Given Line, p. 576. — 536. To Make a Square Equal to two or more Given
Squares, p. 577.— 537. To Make a Circle Equal to two Given Circles, p. 580.
— 53§. To Construct a Square Equal to a Given Rectangle, p. 581. — 539. To
Form a Square Equal to a Given Triangle, p. 582. — 54O. Two Right Lines
being Given, to Find a Third Proportional thereto, p. 582. — 541. Three Right
Lines being Given, to Find a Fourth Proportional thereto, p. 583. — 542. A
Line with Certain Divisions being Given, to Divide Another, Longer or
Shorter, Given Line in the Same Proportion, p. 583. — 543. Between Two
Given Right Lines to Find a Mean Proportional, p. 584.
CONIC SECTIONS.
Art. 544. Definitions, p. 584. — 545. To Find the Axes of the Ellipsis,
p. 585.— 546. To Find the Axis and Base of the Parabola, p. 585.— 547. To
Find the Height, Base, and Transverse Axis of an Hyperbola, p. 585. — 54§.
The Axes being Given, to Find the Foci, and to Describe an Ellipsis with a
String, p. 586. — 549. The Axes being Given, to Describe an Ellipsis with a
Trammel, p. 586. — 55O. To Describe an Ellipsis by Ordinates, p. 588.— 551.
To Describe an Ellipsis by Intersection of Lines, p. 588. — 552. To Describe
an Ellipsis by Intersecting Arcs, p. 590. — 553. To Describe; a Figure Nearly
in the Shape of an Ellipsis by a Pair of Compasses, p. 591.— 554. To Draw
an Oval in the Proportion Seven by Nine, p. 591. — 555. To Draw a Tangent
to an Ellipsis, p. 592. — 556. An Ellipsis with a Tangent Given, to Detect the
Point of Contact, p. 593. — 557. A Diameter of an Ellipsis Given, to Find its
Conjugate, p. 593. — 558. Any Diameter and its Conjugate being Given, to
Ascertain the Two Axes, and thence to Describe the Ellipsis, p. 593. — 559.
To Describe an Ellipsis, whence Axes shall be Proportionate to the Axes of
a Larger or Smaller Given One, p. 594.— 56O. To Describe a Parabola by
Intersection of Lines, p. 594. — 561. To Describe an Hyperbola by Intersec-
tion of Lines, p. 595.
SECTION XVII.— SHADOWS.
Art. 562. The Art of Drawing, p. 596.— 563. The Inclination of the Line
of Shadow, p. 596. — 564. To Find the Line of Shadow on Mouldings and
other Horizontally Straight Projections, p. 597. — 565. To Find the Line of
Shadow Cast by a Shelf, p. 598.— 566. To Find the Shadow Cast by a Shelf
which is Wider at one End than at the Other, p. 599. — 567. To Find the
Shadow of a Shelf having one End Acute or Obtuse Angled, p. 600. — 56§.
To Find the Shadow Cast by an Inclined Shelf, p. 600. — 569. To Find the
Shadow Cast by a Shelf inclined in its Vertical Section either Upward or
Downward, p. 601. — 57O. To Find the Shadow of a Shelf having its Front
Edge or End Curved on the Plan, p. 602. — 571. To Find the Shadow of a
Shelf Curved in the Elevation, p. 602.— 572. To Find the Shadow Cast upon
a Cylindrical Wall by a Projection of any Kind, p. 603.— 573. To Find the
Shadow Cast by a Shelf upon an Inclined Wall. p. 603.— 574. To Find the
Shadow of a Projecting Horizontal Beam, p 604.— 575. To Find the Shadow
624 CONTENTS.
in a Recess, p. 604.— 576. To Find the Shadow in a Recess, when the Face of
the Wall is Inclined, and the Back of the Recess is Vertical, p. 604. — 577.
To Find the Shadow in a Fireplace, p. 605. — 578. To Find the Shadow of a
Moulded Window-Lintel, p. 606. — 579. To Find the Shadow Cast by the
Nosing of a Step, p. 606. — 58O. To Find the Shadow Thrown by a Pedestal
upon Steps, p. 6c6. — 5§1. To Find the Shadow Thrown on a Column by a
Square Abacus, p. 607. — 582. To Find the Shadow Thrown on a Column by
a Circular Abacus, p. 608. — 583. To Find the Shadows on the Capital of a
Column, p. 609.— 584. To Find the Shadow Thrown on a Vertical Wall by a
Column and Entablature Standing in Advance of said Wall, p. 611. — 585*
Shadows on a Cornice, p. 611. — 596. Reflected Light, p. 611.
AMERICAN HOUSE CARPENTER.
APPENDIX.
UNIVERSITY
CONTENTS.
PAGE.
GLOSSARY 627
TABLE OF SQUARES, CUBES, AND ROOTS 638
RULES FOR THE REDUCTION OF DECIMALS 647
TABLE OF CIRCLES 649
TABLE SHOWING THE CAPACTTY OF WELLS, CISTERNS, ETC 653
TABLE OF THE WEIGHTS OF MATERIALS 654
GLOSSARY.
Terms not found here can be found in the lists of definitions in other parts of this book, or in
common dictionaries.
Abacus. — The uppermost member of a capital.
Abattoir. — A slaughter-house.
Abbey. — The residence of an abbot or abbess.
Abutment. — That part of a pier from which the arch springs.
Acanthus. — A plant called in English bear' s-breech. Its leaves are employed
for decorating the Corinthian and the Composite capitals.
Acropolis. — The highest part of a city ; generally the citadel.
Acroteria. — The small pedestals placed on the extremities and apex of a
pediment, originally intended as a base for sculpture.
Aisle. — Passage to and from the pews of a church. In Gothic architecture,
the lean-to wings on the sides of the nave.
Alcove. — Part of a chamber separated by an estrade, or partition of columns.
Recess with seats, etc., in gardens.
Altar. — A pedestal whereon sacrifice was offered. In modern churches, the
area within the railing in front of the pulpit.
Alto-relievo. — High relief ; sculpture projecting from a surface so as to appear
nearly isolated.
Amphitheatre. — A double theatre, employed by the ancients for the exhibi-
tion of gladiatorial fights and other shows.
Ancones. — Trusses employed as an apparent support to a cornice upon the
flanks of the architrave.
Annulet.— A small square moulding used to separate others ; the fillets in
the Doric capital under the ovolo, and those which separate the flutings of col-
umns, are known by this term.
Antce. — A pilaster attached to a wall.
Apiary.— A place for keeping beehives.
Arabesque. — A building after the Arabian style.
Areostyle.— An intercolumniation of from four to five diameters.
Arcade. — A series of arches.
Arch. An arrangement of stones or other material in a curvilinear form, so
as to perform the office of a lintel and carry superincumbent weights.
Architrave.— That part of the entablature which rests upon the capital of a
column, and is beneath the frieze. The casing and mouldings about a door or
window.
Archivolt.—The ceiling of a vault ; the under surface of an arch.
Area.— Superficial measurement. An open space, below the level of the
ground, in front of basement windows.
628 APPENDIX.
Arsenal. — A public establishment for the deposition of arms and warlike
stores.
Astragal. — A small moulding consisting of a half-round with a fillet on each
side.
Attic.— A. low story erected over an order of architecture. A low additional
story immediately under the roof of a building.
Aviary. — A place for keeping and breeding birds.
Balcony. — An open gallery projecting from the front of a building.
Baluster. — A small pillar or pilaster supporting a rail.
Balustrade. — A series of balusters connected by a rail.
Barge-course. — That part of the covering which projects over the gable of a
building.
Base. — The lowest part of a wall, column, etc.
Basement-story. — That which is immediately under the principal story, and
included within the foundation of the building.
Basso-relievo. — Low relief ; sculptured figures projecting from a surface one
half their thickness or less. See Alto-relievo.
Battering. — See Talus.
Battlement. — Indentations on the top of a wall or parapet.
Bay-window. — A window projecting in two or more planes, and not form-
ing the segment of a circle.
Bazaar. — A species of mart or exchange for the sale of various articles of
merchandise.
Bead. — A circular moulding.
Bed-mouldings. — Those mouldings which are between the corona and the
frieze.
Belfry. — That part of the steeple in which the bells are hung ; anciently
called campanile.
Belvedere. — An ornamental turret or observatory commanding a pleasant
prospect.
Bow-window. — A window projecting in curved lines.
Bressummer. — A beam or iron tie supporting a wall over a gateway or other
opening.
Brick-nogging. — The brickwork between studs of partitions.
Buttress. — A projection from a wall to give additional strength.
Cable. — A cylindrical moulding placed in flutes at the lower part of the col-
umn.
Camber. — To give a convexity to the upper surface of a beam.
Campanile. — A tower for the reception of bells, usually, in Italy, separated
from the church.
Canopy. — An ornamental covering over a seat of state.
Cantalivers. — The ends of rafters under a projecting roof. Pieces of wood
or stone supporting the eaves.
Capital. — The uppermost part of a column included between the shaft and
the architrave.
Caravansera. — In the East, a large public building for the reception of trav-
ellers by caravans in the desert.
GLOSSARY. 629
Carpentry. — (From the Latin carpentum, carved wood.) That department
of science and art which treats of the disposition, the construction, and the
relative strength of timber. The first is called descriptive, the second con-
structive, and the last mechanical carpentry.
Caryatides. — Figures of women used instead of columns to support an
entablature.
Casino. — A small country-house.
Castellated. — Built with battlements and turrets in imitation of ancient
castles.
Castle. — A building fortified for military defence. A house with towers,
usually encompassed with walls and moats, and having a donjon, or keep, in
the centre.
Catacombs. — Subterraneous places for burying the dead.
Cathedral. — The principal church of a province or diocese, wherein* the
throne of the archbishop or bishop is placed.
Cavetto. — A concave moulding comprising the quadrant of a circle.
Cemetery. — An edifice or area where the dead are interred.
Cenotaph. — A monument erected to the memory of a person buried in
another place.
Centring. — The temporary woodwork, or framing, whereon any vaulted
work is constructed.
Cesspool. — A well under a drain or pavement to receive the waste water and
sediment.
Chamfer. — The bevelled edge of anything originally right angled.
Chancel. — That part of a Gothic church in which the altar is placed.
Chantry. — A little chapel in ancient churches, with an endowment for one
or more priests to say mass for the relief of souls out of purgatory.
Chapel. — A building for religious worship, erected separately from a church,
and served by a chaplain.
Chaplet. — A moulding carved into beads, olives, etc.
Cincture. — The ring, listel, or fillet, at the top and bottom of a column,
which divides the shaft of the column from its capital and base.
Circus. — A straight, long, narrow building used by the Romans for the ex-
hibition of public spectacles and chariot races. At the present day, a building
enclosing an arena for the exhibition of feats of horsemanship.
Clere-story. — The upper part of the nave of a church above the roofs of the
aisles.
Cloister. — The square space attached to a regular monastery or large church,
having a peristyle or ambulatory around it, covered with a range of buildings.
Coffer-dam. — A case of piling, water-tight, fixed in the bed of a river, for the
purpose of excluding the water while any work, such as a wharf, wall, or the
pier of a bridge, is carried up.
Collar-beam. — A horizontal beam framed between two principal rafters above
the tie-beam.
Colonnade. — A range of columns.
Columbarium. — A pigeon-house.
Column. — A vertical cylindrical support under the entablature of an order.
Common-rafters. — The same as jack-rafters, which see.
630 APPENDIX.
Conduit. — A long, narrow, walled passage underground, for secret com-
munication between different apartments. A canal or pipe for the conveyance
of water.
Conservatory. — A building for preserving curious and rare exotic plants.
Consoles. — The same as ancones, which see.
Contour. — The external lines which bound and terminate a figure.
Convent. — A building for the reception of a society of religious persons.
Coping. — Stones laid on the top of a wall to defend it from the weather.
Corbels. — Stones or timbers fixed in a wall to sustain the timbers of a floor
or roof.
Cornice. — Any moulded projection which crowns or finishes the part to
which it is affixed.
Corona. — That part of a cornice which is between the crown-moulding and
the bed-mouldings.
Cornucopia. — The horn of plenty.
Corridor. — An open gallery or communication to the different apartments of
a house.
Cove. — A concave moulding.
Cripple-rafters. — The short rafters which are spiked to the hip-rafter of a
roof.
Crockets. — In Gothic architecture, the ornaments placed along the angles of
pediments, pinnacles, etc.
Crosettes. — The same as ancones, which see.
Crypt. — The under or hidden part of a building.
Culvert. — An arched channel of masonry or brickwork, built beneath the
bed of a canal for the purpose of conducting water under it. Any arched
channel for water underground.
Cupola. — A small building on the top of a dome.
Curtail-step. — A step with a spiral end, usually the first of the flight.
Cusps. — The pendants of a pointed arch.
Cyma. — An ogee. There are two kinds ; the cyma-recta, having the upper
part concave and the lower convex, and the cyma-reversa, with the upper part
convex and the lower concave.
Dado. — The die, or part between the base and cornice of a pedestal.
Dairy. — An apartment or building for the preservation of milk, and the
manufacture of it into butter, cheese, etc.
Dead-shoar. — A piece of timber or stone stood vertically in brickwork, to
support a superincumbent weight until the brickwork which is to carry it
has set or become hard.
Decastyle. — A building having ten columns in front.
Dentils. — (From the Latin, dentes, teeth.) Small rectangular blocks used in
the bed-mouldings of some of the orders.
Diastyle. — An intercolumniation of three, or, as some say, four diameters.
Die. — That part of a pedestal included between the base and the cornice ; it
is also called a dado.
Dodecastyle. — A building having twelve columns in front.
Donjon. — A massive tower within ancient castles, to which the garrison
might retreat in case of necessity.
GLOSSARY. 631
Dcoks. — A Scotch name given to wooden brick*.
Dormer. — A window placed on the roof of a house, the frame being placed
vertically on the rafters.
Dormitory. — A sleeping-room.
Dovecote. — A building for keeping tarrje pigeons. A columbarium.
Echinus. — The Grecian ovolo.
Elevation. — A geometrical projection drawn on a plane at right angles to
the horizon.
Entablature. — That part of an order which is supported by the columns ;
consisting of the architrave,lrieze, and cornice.
Etistyle.—An intercolumniation of two and a quarter diameters.
Exchange. — A building in which merchants and brokers meet to transact
business.
Extrados. — The exterior curve of an arch.
Facade. — The principal front of any building.
Face-mould. — The pattern for marking the plank out of which hand-railing
is to be cut for stairs, etc.
Facia, or Fascia. — A flat member, like a band or broad fillet.
Falling-mould. — The mould applied to the convex, vertical surface of the
rail-piece, in order to form the back and under surface of the rail, and finish
the squaring.
Festoon. — An ornament representing a wreath of flowers and leaves.
Fillet. — A narrow flat band, listel, or annulet, used for the separation of
one moulding from another, and to give breadth and firmness to the edges of
mouldings.
Flutes.— Upright channels on the shafts of columns.
Flyers. — Steps in a flight ot stairs that are parallel to each other.
Forum. — In ancient architecture a public market ; also, a place where the
common courts were held and law pleadings carried on.
Foundry. — A building in which various metals are cast into moulds or
shapes.
Fneze. — That part of an entablature included between the architrave and
the corn.ice.
•
Gable.— The vertical, triangular piece of wall at the end of a roof, from the
level of the eaves to the summit.
Gain. — A recess made to receive a tenon or tusk.
Gallery.— A common passage to several rooms in an upper story. A long
room for the reception of pictures. A platform raised on columns, pilasters,
or piers.
Girder. -rite principal beam in a floor, for supporting the binding and
other joists, whereby the bearing or length is lessened.
Glyph— A vertical, sunken channel. From their number, those in the
Doric order are called triglyphs.
Granary.— A building for storing grain, especially that intended to be
kept for a considerable time.
632 APPENDIX.
Groin. — The line formed by the intersection of two arches, which cross each
other at any angle.
Gutta. — The small cylindrical pendent ornaments, otherwise called drops,
used in the Doric order under the triglyphs, and also pendent from the mutult
of the cornice.
Gymnasium. — Originally, a place measured out and covered with sand for
the exercise of athletic games ; afterward, spacious buildings devoted to the
mental as well as corporeal instruction of youth.
Hall. — The first large apartment on entering a house. The public room of
a corporate body. A manor-house.
Ham.— A. house or dwelling-place. A street or village : hence Notting-
ham, Buckingham, etc. Hamlet, the diminutive of ham, is a small street or
village.
Helix. — The small volute, or twist, under the abacus in the Corinthian
capital.
Hem. — The projecting spiral fillet of the Ionic capital.
Hexastyle. — A building having six columns in front.
Hip-rafter. — A piece of timber placed at the angle made by two adjacent
inclined roofs.
Homestall. — A mansion-house, or seat in the country.
Hotel, or Hostel. — A large inn or place of public entertainment. A large
house or palace.
Hot-house. — A glass building used in gardening.
Hovel. — An open shed.
Hut. — A small cottage or hovel, generally constructed of earthy materials,
as strong loamy clay, etc.
Impost. — The capital of a pier or pilaster which supports an arch.
Intaglio. — Sculpture in which the subject is hollowed out, so that the im-
pression from it presents the appearance of a bas-relief.
Intercolumniation. — The distance between two columns.
Intrados. — The interior and lower curve of an arch.
Jack-rafters. — Rafters that fill in between the principal rafters of a roof;
called also common-rafters. •
Jail. — A place of legal confinement.
Jambs. — The vertical sides of an aperture.
Joggle-piece. — A post to receive struts.
Joists.— The timbers to which the boards of a floor or the laths of a ceiling
are nailed.
Keep. — The same as donjon, which see.
Key-stone. — The highest central stone of an arch.
Kiln. — A building for the accumulation and retention of heat, in order to
dry or burn certain materials deposited within it.
King-post. — The centre-post in a trussed roof.
Knee. — A convex bend in the back of a hand-rail. See Ramp.
GLOSSARY. 633
Lactarium. — The same as dairy, which see.
Lantern. — A cupola having windows in the sides for lighting an apartment
beneath.
Larmier.— The same as corona, which see.
Lattice. — A reticulated window for the admission of air, rather than light,
as in dairies and cellars.
Lever-boards. — Blind-slats; a set of boards so fastened that they maybe
turned at any angle to admit more or less light, or to lap upon each other so
as to exclude all air or light through apertures.
Lintel. — A piece of timber or stone placed horizontally over a door, win-
dow, or other opening.
Listel. — The same asyf//<?/, which see.
Lobby. — An enclosed space, or passage, communicating with the principal
room or rooms of a house.
Lodge. — A small house near and subordinate to the mansion. A cottage
placed at the gate of the road leading to a mansion.
Loof. — A. small narrow window. Loophole is a term applied to the vertical
series of doors in a warehouse, through which goods are delivered by means
of a crane.
Luffer-boarding. — The same as lever-boards, which see.
Luthetn. — The same as dormer, which see.
Mausoleum. — A sepulchral building — so called from a very celebrated one
erected to the memory of Mausolus, king of Caria, by his wife Artemisia.
Melopa. — The square space in the frieze between the triglyphs of the Doric
order.
Mezzanine.— A story of small height introduced between two of greater
height.
Minaret. — A slender, lofty turret having projecting balconies, common in
Mohammedan countries.
Minster.— A church to which an ecclesiastical fraternity has been or is
attached.
Moat. — An excavated reservoir of water, surrounding a house, castle, or
town.
Modillion. — A projection under the corona of the richer orders, resembling
a bracket.
Module.— The semi-diameter of a column, used by the architect as a meas-
ure by which to proportion the parts of an order.
Monastery.— A building or buildings appropriated to the reception of
monks.
Monopteron. — A circular colonnade supporting a dome without an enclos-
ing wall.
Mosaic.— A mode of representing objects by the inlaying of small cubes of
glass, stone, marble, shells, etc.
Mosque. — A Mohammedan temple or place of worship.
Midlions.—Thz upright posts or bars which divide the lights in a Gothic
window.
Mtiniment-house.—k strong, fire-proof apartment for the keeping and pres-
ervation of evidences, charters, seals, etc., called muniments.
634 APPENDIX.
Museum. — A repository of natural, scientific, and literary curiosities or of
works of art.
Mutule. — A projecting ornament of the Doric cornice supposed to repre-
sent the ends of rafters.
Nave. — The main body of a Gothic church.
Newel. — A post at the starting or landing of a flight of stairs.
Niche. — A cavity or hollow place in a wall for the reception of a statue,
vase, etc.
Nogs. — Wooden bricks.
Nosing. — The rounded and projecting edge of a step in stairs.
Nunnery. — A building or buildings appropriated for the reception of nuns.
Obelisk. — A lofty pillar of a rectangular form.
Octastyle. — A building with eight columns in front.
Odeum. — Among the Greeks, a species of theatre wherein the poets and
musicians rehearsed their compositions previous to the public production of
them.
Ogee. — See cyma.
Orangery. — A gallery or building in a garden or parterre fronting the
south.
Oriel-window. — A large bay or recessed window in a hall, chapel, or other
apartment.
Ovolo. — A convex projecting moulding whose profile is the quadrant of a
circle.
Pagoda. — A temple or place of worship in India.
Palisade. — A fence of pales or stakes driven into the ground.
Parapet. — A small wall of any material for protection on the sides of
bridges, quays, or high buildings.
Pavilion. — A turret or small building generally insulated and comprised
under a single roof.
Pedestal. — A square foundation used to elevate and sustain a column,
statue, etc.
Pediment. — The triangular crowning part of a portico or aperture which
terminates vertically the sloping parts of the roof; this, in Gothic architecture,
is called a gable.
Penitentiary. — A prison for the confinement of criminals whose crimes are
not of a very heinous nature.
Piazza. — A square, open space surrounded by buildings. This term is
often improperly used to denote a portico.
Pier. — A rectangular pillar without any regular base or capital. The up-
right, narrow portions of walls between doors and windows are known by this
term.
Pilaster. — A square pillar, sometimes insulated, but more commonly en-
gaged in a wall, and projecting only a part of its thickness.
Piles. — Large timbers driven into the ground to make a secure foundation
in marshy places, or in the bed of a river.
GLOSSARY. 635
Pitta*.— A column of irregular form, always disengaged, and always deviat-
ing from the proportions of the orders; whence the distinction between a
pillar and a column.
Pinnacle. — A small spire used to ornament Gothic buildings.
Planceer. — The same as soffit, which see.
Plinth. — The lower square member of the base of a column, pedestal or
wall.
Porch. — An exterior appendage to a building, forming a covered approach
to one of its principal doorways.
Portal.— The arch over a door or gate ; the framework of the gate ; the
lesser gate, when there are two of different dimensions at one entrance.
Portcullis.— A strong timber gate to old castles, made to slide up and
down vertically.
Portico. — A colonnade supporting a shelter over a walk, or ambulatory.
Priory. — A building similar in its constitution to a monastery or abbey,
the head whereof was called a prior or prioress.
Prism. — A solid bounded on the sides by parallelograms, and on the ends
by polygonal figures in parallel planes.
Prostyle. — A building with columns in front only.
Purlines. — Those pieces of timber which lie under and at right angles to
the rafters to prevent them from sinking.
Pycnostyle. — An intercolumniation of one and a half diameters.
Pyramid. — A solid body standing on a square, triangle, or potygonal basis
and terminating in a point at the top.
Quarry. — A place whence stones and slates are procured.
Quay. — (Pronounced key.) A bank formed towards the sea or on the side
of a river for free passage, or for the purpose of unloading merchandise.
Quoin. — An external angle. See Rustic quoins.
Rabbet, or Rebate. — A groove or channel in the edge of a board.
Ramp. — A concave bend in the back of a hand-rail.
Rampant arch. — One having abutments of different heights. .
Regula. — The band below the taenia in the Doric order.
Riser. — In stairs, the vertical board forming the front of a step.
Rostrum. — An elevated platform from which a speaker addresses an audi-
ence.
Rotunda. — A circular building.
Rubble-wall. — A wall built of unhewn stone.
Rudenture. — The same as cable, which see.
Rustic quoins. — The stones placed on the external angle of a building, pro-
jecting beyond the face of the wall, and having their edges bevelled.
Rustic-work. — A mode of building masonry wherein the faces of the stones
are left rough, the sides only being wrought smooth where the union of the
stones takes place.
Salon, or Saloon. — A lofty and spacious apartment comprehending the
height of two stories with two tiers of windows.
636 APPENDIX.
Sarcophagus. — A tomb or coffin made of one stone.
Scantling. — The measure to which a piece of timber is to be or has been
cut.
Scarfing. — The joining of two pieces of timber by bolting or nailing trans-
versely together, so that the two appear but one.
Scotia. — The hollow moulding in the base of a column, between the fillets
of the tori.
Scroll. — A carved curvilinear ornament, somewhat resembling in profile
the turnings of a ram's horn.
Sepulchre. — A grave, tomb, or place of interment.
Sewer. — A drain or conduit for carrying off soil or water from any place.
Shaft. — The cylindrical part between the base and the capital of a column.
Shoar. — A piece of timber placed in an oblique direction to support a
building or wall.
Sill. — The horizontal piece of timber at the bottom of framing ; the timber
or stone at the bottom of doors and windows.
Soffit. — The underside of an architrave, corona, etc. The underside of
the heads of doors, windows, etc.
Summer. — The lintel of a door of window ; a beam tenoned into a girder
to support the ends of joists on both sides of it.
Systyle.- — An intercolumniation of two diameters.
Tania. — The fillet which separates the Doric frieze from the architrave.
Talus. — The slope or inclination of a wall, among workmen called bat-
tering.
Terrace. — An area raised before a building, above the level of the ground,
to serve as a walk.
Tesselated pavement. — A curious pavement of mosaic work, composed of
small square stones.
Tetrastyle. — A building having four columns in front.
Thatch. — A covering of straw or reeds used on the roofs of cottages,
barns, etc.
Theatre. — A building appropriated to the representation of dramatic
spectacles.
Tile. — A thin piece or plate of baked clay or other material used for the
external covering of a roof.
Tomb. — A grave, or place for the interment of a human body, including
also any commemorative monument raised over such a place.
Torus. — A moulding of semi-circular profile used in the bases of col-
umns.
Tower. — A lofty building of several stories, round or polygonal.
Transept. — The transverse portion of a cruciform church.
Transom. — The b^am across a double-lighted window ; if the window
have no transom, it is called a clere-story window.
Thread. — That part of a step which is included between the face of its riser
and that of the riser above.
Trellis. — A reticulated framing made of thin bars of wood for screens, win-
dows, etc.
GLOSSARY. 637
\
Tiiglyph.—i:\\e. vertical tablets in the Doric frieze, chamfered on the two
vertical edges, and having two channels in the middle.
Tripod. — A table or seat with three legs.
Trochilus. — The same as scotia, which see.
Truss. — An arrangement of timbers for increasing the resistance to cross-
strains, consisting of a tie, two struts, and a suspending-piece.
Turret. — A small tower, often crowning the angle of a wall, etc.
Tusk. — A short projection under a tenon to increase its strength.
Tympanum. — The naked face of a pediment, included between the level and
the raking mouldings.
Underpinning. — The wall under the ground-sills of a building.
University. — An assemblage of colleges under the supervision of a senate, etc.
Vault. — A concave arched ceiling resting upon two opposite parallel walls.
Venetian-door. — A door having side-lights.
Venetian-window. — A window having three separate apertures.
Veranda. — An awning. An open portico under the extended roof of a
building.
Vestibule. — An apartment which serves as a medium of communication to
another room or series of rooms.
Vestry. — An apartment in a church, or attached to it, for the preservation
of the sacred vestments and utensils.
Villa. — A country-house for the residence of an opulent person.
Vinery. — A house for the cultivation of vines.
Volute. — A spiral scroll, which forms the principal feature of the Ionic and
the Composite capitals.
Vottssoirs. — Arch-stones.
Wainscoting. — Wooden lining of walls, generally in panels.
Water-table. — The stone covering to the projecting foundation or other walls
of a building.
Well. — The space occupied by a flight of stairs. The space left beyond the
ends of the steps is called the well-hole.
Wicket. — A small door made in a gate.
Winders. — In stairs, steps not parallel to each other.
Zophorus. — The same as frieze, which see.
Zystos. — Among the ancients, a portico of unusual length, commonly appro-
priated to gymnastic exercises.
638 APPENDIX.
TABLE OF SQUARES, CUBES. AND ROOTS.
(From Button's Mathematics.)
No.
Square.
Cube.
Sq. Root. jCubeRooJ| No.
Square.
Cube.
Sq. Root.
CubeR-wt.
rr
1
1
1-0000000 1-000000
68
4624
314432
8-2462113
4-081655
2
4
8
1 4142136
1-259921
69
4761
328509
8-3066239
4 1015G6
3
9
27
1-7320508
1-442250
70
4900
343000
8-3666003 4-121285
4
16
64
2-0000000
1-537401
71
5041
357911
8-4261498! 4-140818
5
25
125
2-2360680
1-709976
72
5184
373248
8-4852814J 4-160168
6
36
216
2-4494897
1-817121
73
5329
389017
8-54400371 4-179339
7
49
343
2-6457513
1-912931
74
5476
405224
8-6023253] 4-198336
8
64
512
2-8284271
2-000000
75
5625
421875
8-6602540 4 217163
9
81
729
3-0000000
2-080034
76
5776
433976
8-7177979 4235324
10
100
1000
3-1622777
2-154435
77
5929
456533
8-7749644
4-254321
11
121
1331
3-3166243
2-223330
78
6084
474552
8-8317609! 4-272659
12
144
1728
3-4641016
2-239429
79
6241
493039
8-8881944 4-290840
13
169
2197
36055513 2351335
80
6400
512000
8-9442719! 4-30SS69
14
196
2744
3-74165741 2-410142
81
6561
531441
9-(M)0()000| 4-3*6749
15
225
3375
38729833! 2-466212
82
6724
551358
9-05538511 4-344431
16
256
4096
4'OOOOOOOi 2-519842;
83
6339
571787
9-1104336' 4-362071
17
239
4913
4-12310561 2-571232
84
7056
592704
9-1651514! 4-379519
18
324
5332
4-2426407
2-62074 1 '
85
7225
614125
9-2195445 4-396830
19
361
635J
4-3533989 2-663402!
86
7396
636055
9-2736185 4-414005
20
400
8000
4-4721350 2-714118
87
7569
653503
9-3273791 4-431048
21
441
9261
4-5825757
2-758924
88
7744
681472
9-38J8315J 4-447960
22
484
10643
4-6904158 2-802033
89
7921
704969
9-4339311! 4 "464745
23
529
12167
4-79533151 2-843367
90
8100
729000
9-4363330 4 -58 1405
24
576
13324
4-8989795] 2-881499
91
8281
753571
9-5393920] 4-497941
25
625
15625
5-0000000! 2-924018
92
8464
778688
9-5916630 4514357
26
676
17576
5 -09901 95 j 2-962496
93
8649
804357
9-64365081 4-530055
27
729
19633
5-1961524' 3-000000
94
8836
830534
9-6953597
4 516336
28
784
21952
529151)26 3-036539
95
9025
857375
9-7467943
4-5o2903
29
841
24389
5-3851648 3-072317
96
9216
884736
9-7979590
4-573357
30
900
27000
5-4772256 3-107232
97
9409
912673
9-8483578
4-591701
31
961
29791
556776441 3-14133l|
98
9604
941192
9-8994949
4-610436
32
1024
32763
5-6568542 3-1748021
99
9801
970299
9-9493744
4-626065
33
1089
35937
57445626 3-20753i! 100
10000
1000000
10-0000000
4641589
34
1156
39304
5-83J9519 3233612 101
10201
1030301
10-0498756
4-657009
35
1225
42875
5-9160798 3-271066
102
10404
1061208
10-0995049 41372329
36
1296
46656
6-0000000
3-331927
103
10609
1092727
10- 14839 1GJ 4-687548
37
1369
50653
6-0327625
3-332222
104
10816
1124864
10-1980330! 4-702659
1 38
1444
54872
6-1644140
3-361975
105
11025
1157625
10-2469503 4-717694
«9
1521
59319
6-2449980
3-331211
106
11236
1191016
10-2956301! 4-732623
40
1600
6401)0
6-3245553
3419952
107
11449
1225043
10-3440801] 4-747459
41
1681
68921
6-4031242 3448217
108
11664
1259712
10-3923343' 4-762203
\J
1764
74088
6-4307407 3-476027
109
11831
1295029
10-4403J65' 4-776356
^3
1849
79507
6-5574335
3503398
110
12100
1331000
10-4880335 4-791420
44
1936
85184
6-6332496
3530348
111
12321
1367631
10-535G53S! 4-805895
45
2025
91125
6-7082039
3-556893
112
12544
1404928
10-58300521 4-820284
46
2116
97336
6-7823300
3-533J48
113
12769
1442897
10-6301453 4-834533
47
2209
103423
6-8556546 3-608325
114
12996
1481544
10-0770783
4-848808
48
2304
1 10592
69232032 3634241
115
13225
1520875
107238053
4-862944
49
2401
117649
7-0000000; 3-6593J6
116
13456
1560896
10-7703296
4-87G999
50
2500
125000
7-0710678 3634031
117
13689
1601613
10-816553$
4-890973
51
2601
132651
7-1414284 3-70843J
118
13924
1643032
10-8627805
4-904863
52
2704
140608
7-2111026; 3-732511
119
14161
1685159
10-9087121,
4-913685
53
2809
148877
7-2301099 3-756236
120
14400
1728000
10-9544512
4-93-3424
54
2916
157464
7-3181692 3-779763 121
14641
1771561
11-0000000
4-946087
55
3025
166375
7-4161985! 3-502952 ! 122
14884
1815848
11-0453610
4-959676
56
.3136
175616
7-4833148
3-8258521! 123
15129
1860867
11*0905365
4-973190
57
3219
185193
7-549334 1
3-343501 1! 124
15376
1906624
11-1355287
4-986631
58
3364
195112
7-6157731
3-870877 125
15625
1953125
11-1803399
5-000000
59
3481
205379
7-6311457
3-892996 j 126
15876
2000376
11-2219722
5-013293
60
3600
216001)
7-7459667
3-914868 127
16129
2048383
11-2694277
5-026526
61
3721
226981
7-8102497
3-936497
128
16334
2097152
11-3137085
5-039684
62
3344
233328
7-8740079
3-957891
129
16641
2146689
1 1-35 Vd 167
5-052774
63
3969
250017
7-9372539
3-979057
130
16900
2197000 114017543
5-065797
64
4096
262144
8-0000000
4-000000
131
17161
2248091 11*4455331
5-078753
65
4225
274625
8-0622577
4-020726 132
17424
2299968 11-4891253
5-091643
66
4356
2374%
8-1240334
4-041240 133
17689
2252637 11-53256261 5104469
67
448'J
300763
8-1853523 4-061548|! 134
17956 2406I041 ]i-575«369l 5' 11 7230
TABLE OF SQUARES, CUBES, AND ROOTS.
639
tfo.
Square.
Cube.
Sq. Root. CubeRoot.
No.
Square.
40804
41209
Cube. Sq. Root. CubcRooL
135
136
18225
18496
2460375
2515156
11-6189500 5-129928
11-6619033! 5-142563
202
203
! 8242406
8365427
14-2126704 5-867464
14-24780681 5-877131
137
18769
2571353
11-7016999J 5-155137
204
4I616J 8489664
14-23285691 5-836765
133
19044
2628072
11-7473401
5- 167649
205
42023
8615125! 14-3178211
5-896368
139
19321
2635619
11-7898261
5-180101
206 42 1ST
8741816! 14-3527001
5-905941
140
19600
27-14000
11-83215%
5-192494
207
42849
8369743 14'337494fi
5-915432
141
19881
2803221
11-8743422
5-204328
203
43264
8998912 11-4222051
5-924992
142
20164
2S63283
11-9163753
5-217103
209
43681
9129329 14-4568322
5-934473
143
20449
2924207
11-9582607
5-229321
210
44100
9261000 14-4913767
5-943J22
144
20736
2985934
12-0000000
5-241483
211
44521
9393931] 14-5253390
5-953342
145
21025
3048625
12-0415946
5-253533
212
44944
9528123; 14-5602193! 5-962732
146
21316
3112136
12-0830460J 5-265637
213
45369
9663597J 14-5945195! 5-972093
147
21609
317C>523
12-1243557 5-277632
214
45796
9800344! 146287338! 5-931424
148
21904
3241792
12-1655251
5-289572!
215
46225
99333751 14-6623783! S'%0726
149
22201
3307949
12-2065556
5301459!
216
46656
10077696
14-69693851 6-000000
150
22500
3375000
12-2174487
5-3132931
217
47089
10218313
14-73091991 6-009245
1511 22301
3142951
12-2332057
5-325074
218
47524
10.360232! 14-7648231
6-018462
152i 23104
3511808
12-3238280
5-336803^
219
47961
10503459! 14-7986486
6-027650
153 23409
3531577
12-3693169
5-348481!
220
48400 10648000 14-8323970
6-036811
154 23710
3652264
12-4096736
5-360108!
221
48341, 10793861 14-8660687
6-045943
155 24025 3723375
12-449899. »
5-371685
222
49234 10941048! 14-89 J6044
6-055049
156 24336
37% 416i 12-4399960
5-383213
223
49729
11039567 14-9331845
6-064127
157 24649 3869393
12-5299641
5-394691
224
50176
11239424 149666295 6-073178
153 24964
3944312
12-5698051
5-406120J
225
50625
11390625 15-000000ol 6-082202
159 25281
4019679
12-61)95202
5-417501J
226
51076
11543176 15-0332964 6-091199
160J 25600
4096000
12-6191106
5-428335!
227
51529
11697083 15-05651921 6-100170
161
25921
4173281
12-6385775
5-440122
223
51984
11852352 15-0996639J 6109115
162
26344
4251523
12-7279221
5-451362
229
52441
12008939 15-1327460' 6-118033
163 26569
4330747
12-7671453
5-462556
230
52900
12167000 15-1657509
6-126926
164
26896
4410944
12-8062485
5-473704 231
53361
12326391
151936342
6-135792
165
27225
4492125
12-8452326
5-484807 232
53824
12487168 15-2315462
6-144634
166
87556
4574296! 12-8340987
5-495365 233
54289
12649337 152643375
6-153449
167 27839
4657463 129228480
5-506878!
234
54756
12812904
15-2970585
6-162240
168 23^4
4741632 12-9614814
5-51781H -235
55225
12977875
15-32J70J7
6-171006
169 28561
4826309 13-0000000
5-528775! 236
55696
13144256 153622915 6-179747
170 28'JOO
4913000 13-0331048
5-539658 237
56169
13312053! 15-3J43043I 6-183463
171
2;>2 1 1
5000211 13-0766968
5-550499 233
56644
134812721 154272486 6-197154
172
29581
5083148 13-1143770
5-561293! 239
57121
13651919! 15-4596248 6-205822
173 89929] 5177717
1741 30276 52681)24
13-1529464
13-1909060
5-572055
5 532770
240
241
57600
53031
13324000 15-4919334! 6-214465
139/7521: 155241747 6223084
175
30625
5359375
13-2287565
5-593445'
242
58564
11172433! 15-5563492 6231630
176
30976
5151776 13-2664992 5-604079!
243
59049
14348907
l.V.'.ssif,:;} J6-210J51
177
31329 5515233 13-3041347
31C81 5639752 13-3416541
5-614672!
5-625226
244
245
59536
60025
14526784
14706125
15-6204994 6243300
15-6524753 6-25 7325
179
32041 5735339! 13-3790332
5-635741
246
60516
14836936
15-6343371 6-265327
130
32103 5832000
13-4161071)
5-6462 16
247
61009
15069223
15-7162335; 6-274305
181! 32761] 5921)711
13-4536240 5-656653|| 248
61504
152529J2
15-74801571 6--N2701
182i 33124 6023568
134907376 5-6670511 249
62001
15433249
15-7797333 6-291195
183; 33 kd 6128487
13-5277493! 5-67741 1| 250
62500
15525000
15-8113333! 6-299605
181 33356
6229504
13-5646600! 5-6S7734 251
63001
15313251
15-8429795 6-307994
185 3 1225
6331625
13-6014705J 5-693019
252
63504
16003008
158745079 6316360
18*
34596
613H56
13-63318171 5-70821'.; 253
64009
16194277
•-•737
6-321704
187
3iy;>'.i
653921)3
13-07 17J 13 5-718 17'J -J.M
64516
16337064
15-9373775
6-333026
IMS 35314
6644672
13-7113092 5-728.-,:,l
255 65025
16581375
159687194
6341326
isj 35721
6751269
13-7477271 5-7337J4
256 65536
16777216 160000000
6-349604
190J 36100
6859000
13-78404831 5-1 18391
257 65049
16.)74593
16-0312195
6-357861
191 36181
192 36864
6967871
7077838
13-S202750 5-758966
13-85640651 5-76-.
253 66564
259 67031
17173312 16-06237S1
17373979 16-0934769
6-366097
6374311
193i 372)9
7189057
13-8924140! 5 -77S.»96 200' 67600
l?f>76000
16-1215155
6-382504
194i 37,136
73.H331
13-9283383; 5-783960
261 63121
17779.-.S1 1C,- 1. V.I.I 11
6-390676
195 3302.",
7414875
13-9642400J 5-798390
2621 6S644
17984723 16-1884141
6-398889
196J 38416
7529536
14-0000000! 5-808786
263
69169
18191447 16-2172717
6-406953
197j 33909
7615373
14-0356683 5-818648
264
69696
1SW9711 l:-.2H07t-,<
6-415069
198 392u4
77IWW2
14-0712473 5-S2-U77
265
70225
18609625 16-2738206
(•,•12315-
199) 39H01
7880599
14- 10673(50
5-833272
265
70756 188210961 16-30.).",. >C, I 6-1312;!-
2«XJ' 10000
8000000
14-1421356
5-848035
267
7I-2S9 190341631 16-3401346 6-439277
201 40101
8120601
14-1774469
5-857766
268 71824
19248832, 16 -37070."):) 6-447306
640
APPENDIX.
No. Square.
Cube.
Sq. Root.
CubeRoot.
No. Square.
Cube.
Sq Root. IcubeRnot-
269' 72361
19465109
16-40121951 6-455315
336
1 12896
37933056
18-3303028
6952053
270
72900
19633000
16-43167671 6-463304
337
113569
38272753
183575598
6-958943
271
73441
19902511
16-4620776 6-471274
333
114244
33614472
18-3347763
6-965820
272
73384
20123648
16-4924225! 6-479224
339
114921
38908219
18-4119526
6-972683
273
74529
20346417
16-5227116
6-487154
340
115600
393(UOOO
18-4390889
6-979532
274
75076
20570824
16-5529454
6-495065
341
116281
39651821
184661853
6-986368
275
75625
20796375
16-5331240
6-502357
342
116964
40001638
18-4932420
6-993191
276
76176
21024576
16-6132477
6-510833
343
117649
40353607
18-5202592
7-000000
277
7^729
21253933
16-6433170
6-518634
344
1 18336
40707584
18-5472370
7-006796
278
77234
21484952
16-6733320
6-526519
345
119025
41063625
18-5741756
7-013579
279
77811
21717639
16-7032931
6534335
346
119716
41421736
18-6010752
7-020349
230
78400
21952000
16-7332005
6-542133
347
120409
41781923
18-6279360
7027106
281
78961
22188041
16-7630546
6-549912
343
121104
42144192
18-6547581
7-033850
282
79524
22425763
16-7923556
6-557672
349
121801
42508549
18-8815417
7-040581
233
80089
22665187
16-8226033
6565414
350
122500
42875000
18-7032869
7-047299
284
80656
22906334
16-8522995
6-573139
351
123201
43243551
18-7349940
7-054004
235
81225
23149125
16-8819430
6-530344
352
123904
43614208
18-7616630
7-060697
286
81796
23393656
169115345
6538532
353
124609
43985977
18-7882942
7-067377
287
82369
23639903
16-9410743
6-596202
354
125316
44361864
18-8148877
7-074044
283
82944
23387872
16-9705627
6-603354
355
126025
44738875
18-8414437
7-080699
239
83^.21
24137569
17-0000000
6-611489
356
126736
45118016
18-8679623
7-087341
290
84100, 24389000
17-0293864 6-619106
357
127449
45499293
18-8944436
7-093971
291
84681 24642171
17-0537221 6-626705
358
128164
45882712
18-9208379
7-100588
292
85264 24897083
17-0380075 6634237
359
128881
46268279
18-9472953
7-107194
293 85849 251^3757 17-1172428
6641852
360
129600
46656000
18-9736660
7-113787
294 86 136 25*12184! 17-1464232
6-649400
361
130321
47045381
19-0000000
7-120367
295 87025
25672375
17-1755640
6-656930
352
131044
47437928
19-0262976
7-126936
296 87616
25934336
17-2046505
6-664444
363
131769
47832147
19-0525589
7-133492
297. 83209
26198073
17-2336879
6-671940
354
132496
48228544
19-0787840
7-140037
293 83804
26463592
17-2626765
6-679420
365
133225
48627125
19-1049732
7-146569
299! 89401
26730899
17-2916165
6-686833
366
133956
49027396
19-1311265
7153090
300 90000
27000000
17-3205081
6-694329
367
134689
49430863
19-1572441
7-159599
301
90601
27270901
17-3493516
6-701759
368
135424
49835032
1'.)- 1833261
7-166096
302
91204
27543603
17-3781472
6-709173
369
136161
53243409
192093727
7-172531
303
91809 27818127
17-4068952
6-716570
370
136900
50653000
19-2353341
7-179054
304
92416
28094464J 17-4355953
6-723951
371
137641
51064811
19-2613603
7-185516
305
93025
28372625 17-4642492
6-731316
372
138384
51478848
19-2373015
7-191966
306
93636
23652616 17-4928557! 6-733664
373
139129
51895117
19-3132079
7-198405
307
94249
28934443. 17-5214155
6-745997
374
139876
52313624
19-3390796
7-204832
308
94864
29218112 17-5499288
6-753313
375
140625
52734375
19-3649167
7-211248
309
95481
2951)3629 17-5783953
6-760614
376
141376
53157376
19-3307194
7-217652
310
96100
297910001 17-6068169
6-767899
377
142129
53582633
19-4164878
7-224045
311
96721
30080231
17-6351921
6-775169
378
142884
54010152
19-4422221
7-233427
312
97344
33371328
17-6635217
6-782423
379
143541
54439939
19-4679223
7-236797
313
97969; 30664297
17-6918060
6-789661
380
144400
54872000
19-4935837
7-243156
311
985961 30959144
17-7200451
6-796834
331
145161
55306341
19-5192213
7-249504
315
99225
31255375
17-7482393
6-804092
332
145924
55742968
19-5448203
7-255841
31li
99856
31554496 17-7763383
6-811235
333
146639
56181837
19-5703353
7-262167
317
100489
31855013 17-8044933
6-818462
334
147456
56623104
19-5959179
7-268482
318 101124
32157432 17-8325545
6-825624
335
148225
57066625
19-6214169
7-274786
3191 101761
32461759 17-8605711
6-832771
336
148996
57512456
19-6468327
7-231079
320! 102400
32763000 17-8835438
6-839904
3S7
149769
57960603
19-6723156
7287362
321| 103041
33076161 17-9164729
6847021
338
150544
58411072
19-6977 15f
7-293633
322! 103684
33336248i 17-9443534
6854124
339
151321
53863869 19-7230829
7-299894
323 104329
33598267 17-9722308
6-861212
390
152100
59319000 19-7434177
7-306144
324 104976
34012224 18-0000000
6-868235
391
152831
59776471 19-7737199
7312333
325 105625
34323125 18-0277564
6-875344
392
153664
60236238 19-7989899
7-318611
326
106276
34645976
18-0554701
6-882339
393
154449
60693457 19-8242276
7-324829
327
106929
34965783 18-0831413
6-889419
394
155236 61162984J 19-8494332
7331037
328
107584
35287552
18-1107703
6-896435
395
156025 61623875
19-8746069
7-337234
329
108241
35611239
18-1333571
6903436
396
155816
62099136
19-8997487
7-343420
33C
108900
35937000
18-1659021
6-910423
1 397
157609
62570773
19-9243538
7-349597
331
109561
36264691 18-1934054
6-917396
! 398
158404
63044792
19-9499373
7-355762
332 110224
365943681 18-2208672
6-924356
399
159201
63521199
19-9749844
7361918
333 110832
36926037 18-248237G
6-93130
400
160000
64000000
20-0030001
7-363063
331
11155fi
37259704 18-2756669
6-933232
401
160801
644H1201
200249844
7-374198
335
112225
37595375 18-3030052
6-945150
! 402
161604! 64964308
20 049937*
7-330323
TABLE OF SQUARES, CUBES, AND ROOTS.
641
«0.
Square.
Cube.
Sq. Root.
CubeRoot.
No. Square.
Cube. Sq. Root.
CubeRoot.
403
1024091 65450327
20-0748599
7336437
470! 22J900! 10332300o| 21-6794334
7-774930
404
405
1632 If
164025
F5939264
C643J125
20-099751-2
20-1246118
7-392542
7-398636
471 221841 1044871111 21-7025344
472 222784! 105154048! 21-7255510
7-780490
7-785993
406
164836
66923416) 20-1494417
7 "104721
473| 2237291 105823817 21 -74850*2
7-79148?
0?
165 49
67419143
20-1742410
7-410795
474J 224676 106496424
21-7715411
7-790974
08
1564641 67917312
20-1990099
7-416859
475! 225625
107171875
21-7944947 7-80245-
409
167281 68417929
20-2237484
7-422914
476 226576
107850176
21-8174242) 7-807925
410
168100! 68921000
20-2484567
7-428959
477
227529
108531333
21 84032 J7 7-313389
411
168921 69426i 31
20-2731349
7-434994
47fc
223484
109215352
21-8632111 7-818846
412
169744
69934J 23
20-2977831
7-441019
47S
229441
109902239
21-8850686 7-82429-
413
411
170569
171396
70444997
70957944
20-3224014
20-3469899
7-447034
7-453040
430
431
230400
231361
1 105920UO
111284641
21-908J023 7-829735
21-9317122! 7-H:«16j
415
172225
71178375
20-3715488
7-459035
432
232324
111930168
21-9544934
7-840595
41-
173055
71991296
20-3960781
7-465022
433
233289
112678537
21-9772610
7846013
41?
173389
72511713
20-4205779
7-470999
484
234256
113379J04
22-OOOuOOO
7851424
418 174724
73034632
20-4450483
7-476966
4S5 235225
114084125 y-2'0227155
7-356823
419! 175561
73560059
20-4694895
7-482924
486 236196 114791256
220454077
7-862224
420i 176400
74083000
20-4939015
7-488872
487
237169
115501303) ^-0680765
7-867613
421
177241
74618461
20-5182845
7-494311
488
238144
116214272 22-OJ07220
7-H72994
422
178084
75151448
20-5426336
7-500741
439
239121
116930169 22-1133444
7-878368
423 178929
75636967
20-5669638
7-506661
490
240100
117649000 22 1359435
7-883735
424! 179776
76225024
20-5912603
7-512571
491
241031
118370771! 22-1585193
7-839095
425 180625
76765625
20-6155281
7-518473
492
242064
119095438' 22-1811)730
7-894447
426 181476
77303776
20-6397674
7-524365
493
243049
119823157: 22^035033
7-899792
427t 182329
77854483
20-6639783 7-530248
494
244036
120553784 222251KH
7-905129
428
183184
78402752
20-6881609 7-535122
495
245025! 121287375J 22^35955
7-910460
429
1840 11
78953539
20-7123152 7-541987
4%
246010! 122023936 222710575
7-915733
43li
184900
79507000
20-7364414
7-547842
497
247009
122703473 222934963
7-921099
431
185761
80062991
20-7605395
7-553639
493
243004
1235059 M 22-3159135
7-926403
432 1850*24
80621568!
20-7846097
7-559526
499
249001 124251499) 22^333^79
7-931710
433, 187489
81182737
20-8086520
7565355 500
250000 125000000 22 360o798
7-937005
434 183356
81746504
20-8326667 7571174
501
251001
125751501 22333J293
7-942293
435; 189225
82312875
20-8566536 7'576985
502
252004
126506008 224053505;
7-947574
436 190096 82881856
20-8806130| 7-582786
503
253009 127263527) 22-4270615)
7-952848
437 190969
83453453
20-9045450
7-583579
504
254016 1280^4004 224499443
7953114
433 191844
84027672
20-9284495
7-594363 505 255025 12878702 >i 22-472205 ll
7-963374
439 192721 84604519
20-9523268! 7600133; 506
250036; 129554216 22-4944433
7-968627
440 193600 85184000
20-9761770
7-005905 507
257049 1303233431 22 516651)5
7-973373
441! 194481 8576)121
442; 195364 86350388
21-0000000 7-611663 508
21-0237960) 7-617412 509
253064 131090512! 22 5333553'
259031 131872229) 225510243
7-9791 U
7-934344
443! 196249
86938307
21-0475652! 7-623152 510! 2601001 132651000) 22'5331795|
7-989570
444 197136 87528334
21-0713075! 7-023384 511
261121 133432:131! 22-6053091!
7-994788
445 198025 88121 125 !
21-0950231
7-63460? 512
202144 134217728 22-0274 17o!
8000000
446! 1939161 88716536
21-1187121
7010321 513
263169 13500569?) 22-6195033
8-00521 *
447
199809; 89314623
21-1423745 7010027)! 514
264190 1357:Mi74l
220; 15531)
448
2007041 89915392
21-1660105 7-051725! 515
265225 136590875
22-6936 1 1 4
8-0155'JJ
449 2016011 90518349
21-1896201
7657414 510
266256 137333096
227156334
8-020779
450 202500! 91125000
212132034 7-663094 517
2672891 13318*413
22-7376340! » 02595?
i:-l 203401 91733851
21-2367606 7-663760 5 Is
263324 138991832
22-75J6134)
8-031129
452 204304 92345403
21-2502916 7-071430- 519
269351S 139?'.K!.-.'
22-7815715
8-03529/
453 305209) 92959677
21-2837967; 7'080036 520
•270400 1406U8000
22-8u350a5
M-041451
454 206116 93576664
21-3072758' 7'685733 521
271441 141420761 228254244
8-040003
4531 207025 94196375
21-3307290 7'69137'2 52'2
•272 4^1 142236648
22-847311*3
8-051748
456 21)7936 94818816
21-3541565; 7 6J7002 523
•273529 1 UUV.OO;
22-3691933
8 056336
457! 208349 95443993
21-3775533 7'702025 524
274576 143377824
22-89104 631
8062018
458; 209764
96071912
21-4009340 7-708239 5J5
2750-25 1417.131,5
•2-2-9128785
8-067143
459i 210581
96702579!
21-4242353 7713S15 526
276676 145531570
22-93 10 ^99
8-07-2262
460 211600
97336000
2 1-4 176 UK; 7-719413 5-27
277729 145363183
22-95.i4S.j6
8077374
461 212521
97972181
21-4709 UK") ?-725:>32 528
278784 147197952
22-973^500
8-082430
462 213444
ys.il 1123
21-4941853 7 730014 529
279341 148035339
•2300JOJOO,
463 214369
99252347
21-51743J8 77361HH -,30
280900 148877000
23-0217.^9
809267*
464 215296
99897344
21-5406592 7711753 531
281961J 149721291
•23(1134372
8-097759
465! 216225 1005! 1625
21-56335871 7'7473lir 532
283024! 150558763
23-065 1 -2 J 2
8-102339
466; 217156) 101194696
21-5370331
7-752861 533
284089 15 14 1'9 i:i7
23080792.3
8-107913
46? 21&J39J 101847563
21-6101828
7-758402 531
285156 152273304
23 1084400
B'ltsfctiO
46S 2190241 102503232
21-0333077
7-703.-30 5!55
286225 153130375
23 13, "0070
8-118041
469 219961 10)161709
21-6564078
7-769462 536
287296 153990656 23 1516738
642
APPENDIX.
No.
Square.
Cub*.
Sq. Root.
CuheRoot-' No. Squar,;.
Cube. 1 S.]. Rcu.
CutijKoot.
537
2333691 154854153 23-1732605' 8-123145J 604
3-4816
220348864
245764115
8 453026
£33
239444| 155720872; 23-1948^70 8-133187| 605
3;. 6025
221445125
24-5967478 8-457691
539
2935211 156590819; 23-2163735 8-1332231
606
3<>7236
222545016
24-6170673! 8-462318
540 2916001 157464030; 23-23790011 8-143253
607
368449
223648543
24-63737001 8-467001)
541 292631* 158340421; 23-2594067. 8-148276
608
369664
224755712
24-6576560
8-471647
542 293764', 1592201)881 23-2H03935! 8-153294
609
370881! 225866529
24-6779254
8-476289
543| 294349' 160103007 23-3023604! 8-158305
610
372100] 226981000
246981781
8-480926
544 295936
160939184 23-3233076 8-163310! 611
373321
228099131
24-7184142
8-485558
545 297025
161873625 23-3452351
8-168309 612
374554
229220928
24-7386338
8490185
546 298116
162771336J 23-3665429
8-173302 613
375769
230346397
24-7538358
8-494806
547J 299209
163667323; 23-3330311
8-178239
614
376996
231475544
24-7790234
8-499423
548, 300304
1645665921 23-4093998
8-183269
615
373225
232608375
24-7991935
8-504035
5 19| 301401
.550 302500
165469149
166375000
23-4307490
23-4520788
8-188244
8-193213
616
617
379456
330689
2337448961 24-8193473
234885113 24-8394347
8-508642
8-513243
551
3J3601
1672841511 23-4733392
8-198175) 618
331924
236029032! 24-8596058
8-517840
552
304704! 163196608
23-4946802
8-2031321 619
333161
237176659 24-8797106
8-522432
553 3J5809
16911237.7
23-5159520
8-203032 620
334400
238328UOO] 24-8997992! 8-527019
554 306916
170031464
235372046
8-213027 621
385641
239433061 24-91987161 8-531601
555 303025
170953375
235534333
8-217966 622
386884
240641848
24-9399278 8-535178
556 309136
171879616
23-5/96522
8-222893J 623
38^129
241804367
24-9591679 8-540750
557
310249
172308693
23-6003474
8-227825
i 624
339376
242970624
24-9799920 8'5453l7
558
311364
173741112
23 6220236
8-232746
625 ! 390625
244140625
25-0000000 8-549880
559 312431
174676879! 236431808] 8-237661
626 1 391876
245314376
25-01-99920 8-554437
560 313600
175516000 23-6543191
8-242571
627 393129
246491833
25-0399681
8-553990
561 314721
176553481
23-6854336
8-247474
623 394334
247673152
25-0599282
8-563538
562! 315344
177504328
23-7065392
8-252371
629 395641
248853189
25-0793724
8-568081
563 316969
178453547 23-7276210
8-257263
630
396900
250047000
25-0993003
8-572619
564 318096
17941)6144 23-7481x312 8^62149
631
393161
251239591
25-1197134
8577152
565! 319225
180362125 23769728.1
8-267029! 632
3J9424
252435968
25-1398102
8-581681
566 320356
1813214% 23-7J07515 8-2719041! 633
40068^
253638137
25-1594913
8-536205
567 321489
182284263
23-81176 18| 8-2767731' 634
401956
254810104 25-1793566 8-530724
568
322624
183250432
23-83275:;6: 8-281635! 635
403225
256047375 25-1992C63J 8'595233
569
323761 184220009
33*537409 8-236493 636
404 »96
257259456
25-2190404 8'599748
570
324900 185193000
23-8746728! 8-2 9 1344
637
405769
258474853
25-2333539 8'f>042V2
571
326011! 186169411
238.156063 8-2J6190
638
407044
239694072
25-25366191 8-608753
572
327184
187149248
23-9165215 8-3J1030
639! 408321
260917119
25 2731493 8-613248
573
328329
183132517
23-9374184 8-3J5865 640 409630
262144000
25.2932213! 8-617739
574
3294761 189119224
23-9532 J71 8-310694
641 410381
263374721
25-3179778J 8'622225
575
330625 190109375
239791576 8-315517
642J 412164
264609283
•25-3377189 8-626706
576 331776 191102976
24-0000030 8-3203M31 643
413419 265847707
25-35741471 8-631183
577! 332929
192103033
24-0208213
8-325147! 64-1
414736
267089984
25-3771551
8-635655
578 334034
579! 335241
193100552
194104539
21-0416306
24-0524183
8-329954 645 416025
8-33475J;; 616: 417316
268336125
269586136
25-3968502 8.640123
25-41653.il 8-6445S5
580| 336400 195112000
24-0331891 8-339551 ! 647 418509
270340023 25 -4 3,5 1947
8-649044
5811 337581, 196122941
24-1039416 8-344341 648; 419904
272097792 25 '45581411 8-653497
582) 333724! 197137368
24-1246762! 8-349126! 649 421201
2733594491 254751784: 8-657946
5331 339839
193155237
24-1453929 8-353905i 650i 422500
274625000 25'4950976i 8-662391
534J 341056 199176704
24-1660919
8-353678 65li 423301
275894451 25-5147016: 8-666331
535 342225! 200201625
24-18(57732
83534471 652! 425104
2771(57808
25-53429071 8-671266
586 343395 201230055
242)74359
8-333209! 653 425409! 278445077
25-5533647! 8 675697
587 3445691 202262003
24-2230829
8-372967: 654 427716! 2*79726264! 25'5734237l 8-680124
538 3457441 203297472
242487113 8-377719: 655 4290251 281011375! 25'5929678; 8-68454f
539J 346921 204335469
24-2593222 8-332465
1 656 43,336 282300416 25*6124969 8-683963
590 343100, 20537'JOOOJ 242399156 8-337206 657i 431619! 2835933931 25-6320112 8693376
591 349281 206425071 24-3104916 8-3919421' 658' 432964 234890312 25'6515107 8-697781
*92' 350464: 207474683 213310501 8-3J6673! 659< 4342811 236191179! 25'6709953 8-702188
593! 3516491 208527357
24-3515913 8-4013981 660i 435600! 287496000! 25-6904652; 8-706538
594i 352836 209584584
24-3721152; 840611$,! 661| 436921
288804781 25*7099203 8-710988
595
354025 210644875
24-3926218! 8-410333J 6621 438244
290117528 25-7293607! 8-715373
596
355216! 211703731
24-4131112 8-415542! 663 439569
291434247 25'7487854 8-719760
59?
356409 212776172
24-4335834 8-420246! 664 440896
292754944: 25'7681975 8724141
59$
357604 213347192
24-45403851 8-424945 665 442225
294079625! 25'7875939! 8-728518
59£
353801! 21492179ii
21.4744765 8-429633! 666 443556
295408296; 25-8069758' 8-732392
601
360001
216000001
244948974
8-4343271; 667 444889
296740963! 258263431; 8-737260
GO]
361201
217081801
24-5153012
8-439010! i 668 446224 298077632; 25 84561J6J! 8741625
60S
362404 21816720?
2 4 -5356 -HI
8-443688!' 669, 447561 299418309 25-8(5503431 8-745985
605
363609; 21925622?
24-5560532
8-448360J 670 448900 300763000! 25 "884 35 821 8-753340
TABLE OF SQUARES, CUBES, AND ROOTS.
643
No,
Square.
Cube. | Sq. Root. CubeRoot.
No.
Square.
Cube.
Sq. Ro<t. CubeRool.
671
450241
302111711
25-90366771 8-754691
738
544644 401947*72 27-16615541 9-036880
672
451534
303464448
25-9229623! 8-759033
739
546121
403533419 27-1845544' 9 040965
673
674
452929
454276
304821217
306182024
25-y422435| 8-763331
25-9615100 8-767719
740
741
547600
549031
405224COU 27-2029410! 9'04504*
4U6369021 27-2213152 9-049114
675
455625
307546375
25-98076211 8-7720531 742
550564
408518488 27-23J6769 9'053l83
676
45b'97<i
308915776
26-0000000' 8-776333' 743! 552J49
410172407 27-2530263' 9-057245
677
453329
310288733
26-0192237 878J708 744 553536
411830784 27-2763634
9061310
678
459684 311665752
26-03813311 8-7850301 745! 555025
413493625; 27-2946881
679
630
461041
462401.
313046839
314432000
26.0576284J 8-789347
26-07630%! 8-793659
746 5565 lb
747| 5J80(,9
415160936! 27-31300L6
416332723! 27-33130i)7
9-073473
681
463761
315821241
26-0959767 8-797968
74UJ 5J9504
4185U8992
27-34y5337
y0775*0
632
46-H24
bl7214568
26-1151297 8-802272
749 561001
42Jl8974y
273670644
683
466489
313611987
26-13426871 8-8U6572
1 750 5625 JO
42187.XXX)
27-336127J
y-085bOb
684
4678561 320013504
26-1533937
8-810868!) 7511 564o01
4*3564751
27-40437y2
9-08^63j
685
469225
321419125
26-1725047
8-815160
752 565504
4*5^59008
27-4226134
y-0j36<2
686! 470596
322828356
26-1916017
8-819447
753
o.j/uoy
426957777
27-4403455
9-Oi)770l
<>87i 471961)
688J 473344
3*4242703
32566(;672
262106848
26-2297541
8-823731
8-823010
754
755
568516
570025
4*^661064
430J68375
2745906o4
27-47/2633
9-l0172b
9-105740
609
474721
327082769
26-2483095
8-832285
756
571536
432084215
27-4^54542 9-109767
690
476100
323509UGO
26-*673511
8-836556
757
573U49
433798093
9-113702
691
477431
329939371
26-2868789
8-840823
758
574564
43551951*
2/-5317yy8
y-l!77»3
692
478361
331373888
26-3053929 8-845085
759
57608 1
437*45479
275499546
9-121801
693
480249
33*2812557
26-3248932! b'Ws>344ll 760
577600
438976000
2 T -5680975
y- 125805
<94
481636
334255384
26-34387^7
8-853598 761
57^121
440711081
27'53b2234
9-12y8b6
695
483J25
335702U75
26-3628527
8-857849 762
530644
442450728
27-6043475
y-13380b
696
484416
337153536
26-3318119
8-862095 763
532469
444194947
276224546
y- 1377^7
697
485809
338608373
26-4007576
8.866337 764
533696
4459437^4
9-441787
698
48720-i
3400GS392
26-4196896
8-8705761 765
585225
447697425
^7 'Ojt^o33-i
9- 44577 1
699
488601
341532099
26-4386081
8-874810 766
586756
44*4550b6
27 b7o705b
9-l4975o
700
49UL'00
343000000
26-4575131
8-879040 767
533^09
45121.663
*7'694764o
y-45o737
701
491401
344472101
*6-4764046
8-883266
768
589824
452984332
*7-7l23i*y
y-.'5/Vi^
702
492i04
345948408
26-4952826
8-887483
769
591361
454756609
27-730849*
y-lblbo/
703 491209
347428927
26-5141472
8-891706
770
592900
456533000
27-7480739
9-165656
704
495616
348913664
26 5329983
8-895920
771
594441
45334401 1
27-7663860
9'169b2.i
705
497025
b5u40~625
26-5518361
8-900130
772
5J5984
4b009b648
9-17353J
706
498436
351895316
26-57C6605
8-904337
773
597529
46183i)9i7
278023775
9-17/014
707
499849
353393243
26-5394716
8-908539
774
599076
463684824
27 "8200555
y 401500
708
501264
354894912
26-6082694
8-912737
775
600625
465481375
27 -0 33021 8
y 105453
709
502681
356400829
266270539
8-916931
776
602176
4b7*8e5/6
27 o5b776o
9 18940^
710
504100
357911000
26-6453*52
8-921121
.777
603729
461J097433
27-8747197
y I9b347
711
5U5521
35942543!
266645333
8-925308
778
605284
47U9l09;)2
27'0b26ol4
'j-197~9c.
712
5t6944! 360944128
266833231! 8-929490
779
606841
47V729139
27-9ib57J5
V*0l22lj
713
500369 362467097
26-7020598
8-9b3669
780
6034 b
4/4552UOO
27-9284801
>j-2Ujlt>«
714
5097^6: 353994344
26-7207784
8-937843
781
60^961
47t 37^541
27-y4b3772
y^o^o-Jb
715
511225! 365525375
5T -739483 J
8-942014
782
6115^4
47o^ll768
27-9642629
y-*i3o~o
716
512656 367061696
26-7581763
8-946181
783
613089
40X04060V
27'i»821b7*! y-*ibi)Jv,
717
514089 368601813
26-7763557! 8-950344
; 784
614656
481890^04
23-OCOOObO
y-*2O8/o
718
515524 370146232
26-7955220 8-9545o3|: 785
616225
40o/b66*5
28 Ol7o5l5
>j-»2l7v-l
719
516%l| 37169495U
26-81447.-)!
8-958653 706
617796
4855o'/b5o
28 "0356 yl 5
y-*2o7^/
720
518400; 373248UOO
26-8328157
8-962809! 787
61936'J
407443-lob
28-0535*00
•.'•_. 'Jo i'..
721
519341 37181)5361
26-8514432
8-966957
i 788
620944
48X)3o3i72
28-071 33 //
y-2ot5^6
722 521234 37t.3,'.70v-
26-8700577 8-971101
789
62*521 4yil6906'j
28-089 143^
y-24o4b;>
723 522729 377933i>r>7
26-8886593 8-975241
i 790
6*4100 4y3b3yOoO
23'106y38b
9-*14b3o
724! 524176 379503424
26-9072481
8-979377
i 7<J1
625631
494940671
if8'12472*2
y-*48*b4
725 525625 381078125
26-9258240
8-983509
i 792
627264
49b'/'yi)oM^
23'l42494b
y-2o2i3t
726 527076 382657176
*6-9 143872 8'9876:>7 793
628849 4906/7*57
23-1602557
y-^obo2*
727 528529 38424053«
2-3-9629375 8-99176^ 7J<
63Ji:-.t} o005b6404
28'170bOJb
9-*j99ll
728! 529984' 38582835:.
.6-98 14751 8-995883|: ?y5
6320*5 50*459375
28- 495 i 444
y-*6.uy<
729 531441 3^7420489
27-000(K)00 y-COOOOb
?y6
6336 1<)
5D4353336
20-2134720
y-2b7bOv
730 532900 38^0171-00
27-0185122 9-004113
79.
635*0'J
50626l57«>
28-2311004
'.I -. l-'-'v
731 534361; 39C6l7s.ll
270370117: 9-008223
I 798
636oOi
5084bu5»*
28'248893o
y-^ij4oo
7321 535324 392223168
:.7-c<:>.V;985 9-012329
1 7yy
638401
5iot8*3yy
28*005801
9-i7y3oc
733! 5272*9 39383^837
27-G73V727 9-016431
> 800
64000C
512000000
28-204271*
y-2o3i7o
734! 53/7;»6 3.i54469i»!
27-0924341 'J -02052'.
801
641604J 513922401
28-3019434
y-2a 7o-*-»
735 : 540^25 3J7065375
27.1108834 9-021621
802
643204 51534l6bO
20-3 11.* oi.)
9-290.0*
736 541*i»> 39'0,s(su;>(i 271293199 9-0*a7l5
i 803
644809 51778162/
28-o3/*54b
9-*'j4<b<
737, f>431t.9 40031555:> 27 147743.* 9-032802
804] 64641b
5197181b4
*tt-354oy30
y-2y8ti*i
644
APPENDIX.
No.
Square.
Cube.
Sq. Root.
CubeRootJ
No.
Square.) Cube.
1
Sq. Root CubeKoot.
805
6430251 52166J125
2837*5219 9-302477
872| 761)334
663.)M348| 29-5296461
9-553712
806
649636 52360G616:
283901391
9-306323
873 762129
665338617! 29-5465734
9-557363
807
651249 525557943;
28-4077454
9-310175
874
763376
667627624 29-5634910
9-561011
808
652864
527514112
28 4253408
9-3140191
875 765625
669921875
29-5303989
9-564656
809
654481
529475129
28-4429253
9-317860;
876
767376
<>?2221376
29-59729721 9-568298
810
656100
531441000
28-4604989
9-321697 877
769129
674526133
29-6141853 9-571938
811
657721
533411731
28-4780617
9-325532
878 770884
676836152
29-63106481 9-575574
812
659344
535387328
23-4956137
9-32J363
879 772641
679151439
29-6479342, 9-579208
813
660969
537367797
23-5131549
9-333192
880 774400
681472000
29-6647939! 9-582840
814
662596
539353144
23-5306852
9-347017
b81 776 Hi 1
683797341
29-6816442 9-:86468
815 664225
541343375
23-5482048
93408391
832 777924
686128968
29.6984843
9-590094
816
665356
543338496
23-5657137
9-344657!
883 779689
688465387
29-7153159 9-593717
81?
6l>7489
545333513
28-5832119
9-3:8473
834 781456
690807104
29-7321375' 9-597337
818
669124
547343432
23-6006993
9-352286
8b5j 783225
693154125
29-7439496! 9-600955
819
670761
549353259
28-6181760
9-356095
886 784996
695506456
29-7657521 9-604570
820
672400
551368000
28-6356421
9-359902!
88?
786769
697864103
29-7825452
9-603182
821
674041
553387661
28-6530976
9363705
838
788544
700227072] 29-7993239
9-61*791
822
675684
555412248
28-6705424
9367505:
889
790321
702595369
29-816103:)
9-6153'J'r
823
677329
557441767
23-63797G6
9-371302
890
71J2100
704969000 29-8323678
9-619002
824
678976
559476224
23-7054002
93750961
891
793881
707347971 29-8496231
9-622603
825
680625
561515625
28-7228132
9-373387
892
795664
709732288 29-8663690
9-626202
826
632276
563559976
28-740215?
9-382675
893
797449
712121957 29-8831050
9-629797
827
633429
565609283
28-7576077
9-336460
894
799236
714516934
29-8998328
9-633391
823
685584
567663552
28-7749891
9-390242
895
801025
716917375 29-9165506
9-636981
829
687241
569722789
28-7923601
9-394021!
896
802816
719323136 2l ^332591
9-640563
830
6839 >0
571787000
28-8097206
9-3. 7796!
897
804609
721734273
29-9499583
9-644154
831
690561
5?385619l| 28-8270706 9-401569
898
806404
724150792
29-9666481
9-647737
832
833
69i224
693889
5?5930368| 28-8444102; 9-405339
5780095371 28-86173941 9-409105
899
900
808201 726572699
810000| 729000000
29-98332-37 9-65131?
30-0000000! 9-654894
834
695556
530093704
28-8790582! 9-412869
901
811801 731432701
30-0166620 9-658468
835 697225
582182875
28-8963666! 9-416630
902
813604! 733870808
30-0333148 9-662040
836: 698896
584277056
28-9136646 9-420337
903
815409
73631432?
30-0499584 9-665610
837 700569
586376253
28-9309523
9-424142
904
817216
738763264
30-0665923 9-669176
838 i 702244
5^8480472
28-948229?
9-4278J4
905
819025
741217625
30-0332179
9-672740
839 703921
590589719; 23-965496?
9-431642
906
820836
74367741<i
30-0998333
9-676302
840! 705600
592704000 28-9827535 9-435338
90?
822649
746142643
30-1164407
9-679860
841 707281
594823321
29-OOOOOOC
9-439131
908
824464| 748613312
30-1330383
9-f>83417
842J 708^64
596947688
29-0172362
9-442370
909
826281
751089429
30-149626'J
9-036970
843 710649
599077107
29-0344622
9-446607
910
828100
753571000
30-1662063
9-690521
844j 712336! 6012 1 1584
29-0516781
9-450341
911
829921
756058031
30-1827765
9-694069
8451 7140251 603351125
29-0688837
9-454072
912
831744
758550528
30-1993377
9-697615
8461 715716J 605495736
29-0860791
9-457800
913
833S6S
761048497
30-2158393
9-701158
847 717409 607645123 29-1032644
9-461525
914
835396] 763551944
30-232432b
9-704699
84fc
719104
609300192) 29-1204396
9-465247
915
8372251 766060875
30-248966S
9-708237
84£
720801
611960049 29-137604f
9-463966
916
839056 768575-296
30-26549 It
9-711772
85C
> 72250C
614125000 29-154759=
9-47*682
917
840889 771095213
30-282007L
9-715305
851! 724201
6162950511 29- 17 19042
t 9-476396
918
842724
773620632
3J-2(J8514fc
9-718835
852! 725904
618470208! 29 1890390 9-480106
919
8445611 7761515511
30-3150 12b
9-722363
853 72760'j
6206504771 29-2U61637J 9-483814
92u
84640C
1 773683000
30-3315016
9-725388
854
7293 If
622835364 29-2232784 9'48?518
921
848241
781229961
30-347981^
9-729411
85f
7310^'
625026375 29-2403830 9-491220
922J 850084 78377744ti
3J-364452L
9732931
856
, 73273f
627222016 29-2574777 9-494919
923 851929 786330467
303809151
9-736448
85'
' 73444i
629422793J 29-2745623! 9-498615
924
853776: 78388902-4
30-3J7363?
9-73J963
85£
J 736164
631623712! 29'2916370| 9-5U23J8
925
855625! 791453125
30-4138127
9-743476
85i
> 737881
633339779 29-3087018! 9-505998
926
857476 79402277f
3J-43J2481) 9-746986
860 73960C
6360560(X)| 29 3257566 9-509685 927
859329 796597982
3J-4466747 9-750493
861 74132]
6382773811 29-3423015J 9-513370
928
861184
79917875.
3U-4630924I 9-753J98
862 743044
640503928 29-3598365 9517051
92'J
fc63041
801765031
30-47y5013| 9-757500
863 74476i
6427356471 29-3768616 9-520730
931)
864900 804357000 30-4i/59014| 9-761000
8641 746496! 644972544! 29-3933769 9-524406
931
866761' 806954491 30-5122926J 9-76449?
865 748225 6472146251 29-4103823 9-528079
932
868624. 8095575681 30-5285750 9-767992
866 749956 649461896 29-4278779 9-531750
932
87J4*-»j 812i6G237| 30-545048? 9-771484
867 75168i
> 651714363 29-4448637 9-535417
934
872356; 814780504 30-5:5 14 136! 9-?74i>74
868; 75342^
[ 653972032' 29-4618397 9-539082
933
874225! 817400375
30-67776911 9-778462
869! 75516]
656234909 29-4788059 9-542744
93fc
876096: 820025y5f
30-5;;4Ii71 978194?
87C1 75690(
) 658503000 29-4957624 9 "5 46403
937
877969: 82265615;
3D-610455?; 9 785429
87 1\ 75864
660776311! 29-51270'Jli 9-55UU5!)
933 H79H44 825293675!
30-6267857 9-736908
TABLE OF SQUARES, CUBES, AND ROOTS.
645
No.
939
Square.
881721
Cube. | Sq. Root.
CubeRoot.
No.
Square.
Cube.
Sq. Root.
CabeRooiJ
827936019: 30-6431069
9-792386
970
940900
912673000
3H448230J 9-8389831
940
883600
830584000! 33-6594194
9-795861
971
942341
915498611
31-160872'J
9-902333
941
885481
8332376211 30-6757233
9-799334
972
944784
918330048
31-1769145
9-905782
942
837364
835896888 30-6920185
9-802804
973
946729
921167317| 31-1929479
9-909173
943
889249
83856 1807J 30-7083051
9-806271
974
948676
924010424! 31-2039731
9-912571
944
8U1136
8412323341 30-7245830
9-809736
975
950625
926859375J 31-2249900
9-915962
945
893025
843908625 30-7408523
9-813199
976
952576
929714176
31-2409987
9-919351
946
894916
846590536 30-7571130
9-816659
977
954529
932574833
31-2569992! 9-922733
947
896809
849278123 30-7733651
9-820117
978
956484
935441352
31-27299151 9-926122
948
898704
851971392! 30-7896086
9-823572
979
958141 938313739
31-288.^757
9-929504
949
900601
8546703491 30-8058436
9-827025
980
960400
941192000
31-3049517
9-9328S4
950
90:2500
8573750001 30-82207001 9-830476
981
962361 944076141
31-3209195 9-936261
951
904401
860085351 308382879 9'833924
982
964324! 946966168
31-3368792 9-939636
952
906304
862801408: 30-85449721 9-837369
983
966289 949862087
31-3528308
9943009
953
908209
865523177 30-870698l' 9-840813
984
968256! 952763904
31-36877431 9-946380
954
910116
868250664 30-88689041 9'844254
985
970225 955671625; 31-3847097! 9-949748
K55
912025
870983875 30-9030743, 9-847692
986! 972196
958535256
31-400H369) 9-953114
956
913936
873722816 30-9192497J 9-851128
987; 974169
961504803
31-4165561 9-956477
957
915849
876467493; 30-9354166! 9-854562
2881 9761441 9644302"2
31-4324673! 9-959839
958
917764
879217912 30-95 15751J 9-857993
989
9781211 967361669
31-4483704 9-963198
959
919681
881974079 30-9f>77251| 9-861422
990
980100
970299000
31-4642654 9-96655.l>
960
921SOO
884736000^ 30-9^36668 9-864848
991 982081
973242271
31-4801525 9-969909
961
923521
887503681 31-OOOOOOOi 9-868272
992 984064
976191488
31-4960315! 9-973262
962
925444
NW277128 31-01612481 9-871694
993i 986049
979146657
31-5119025
9-976612
963
927369
;<<.»:J056347 3l-0322413i 9875113
994 988036
982107784
31-5277655
9-979960
964
929296
S;).r)841344 31-0483494J 9-878530
995 990025
985074875
31-5436206
9-983305
965
931225
898632125 31-0644491 9-881945
9%
992016
988047936
31-5594677
9-986649
966
933156
90M28696 21-08054051 9-885357
9971 994009
991026973J 31-5753068
9-989990
967
935089
104231063 31-0966236 9-888767
998 996004
994011992
31-5911330
9-9933291
968
937024
907039232 31-1126984 9-892175
999
998001
997002999
31-6069613
9-996666,
969 938961
909853209 31-1287648! 9-895580 1000
1000000 1000000000
31 -6227766 lO-OOOOOOJ
The following rules are for finding the squares, cubes, and roots of num-
bers exceeding 1000.
To find the square of any number divisible without a remainder. Rule. — Di-
vide the given number by such a number from the foregoing table as will
divide it without a remainder ; then the square of the quotient, multiplied by
the square of the number found in the table, will give the answer.
Examfle.—Whzt is the square of 2000 ? 2000, divided by 1000. a number
found in the table, gives a quotient of 2, the square of which is 4, and the
square of 1000 is 1,000,000, therefore :
4 x 1,000,000 = 4,000,000: the Ans.
Another Example. — What is the square of 1230? 1230, being divided by
123, the quotient will be 10, the square of which is 100, and the square of 123
is 15,129, therefore :
100 x 15,129= 1,512,900: the Ans.
To find the square of any number not divisible without a remainder. Rule. —
Add together the squares of such two adjoining numbers from the table as
shall together equal the given number, and multiply the sum by 2 ; then this
product, less i, will be the answer.
Example. — What is the square of 1487 ? The adjoining numbers, 743 and
744, added together, equal the given number, 1487, and the square of 743 =
552,049, the square of 744 = 553,536, and these added = 1,105,585, therefore :
1,105,585 x 2 — 2,211,170 — i = 2,211,169 : the Ans.
To find the ciibc of any number divisible without a remainder. Ritle. — Divide
the given number by such a number from the foregoing table as will divide
646 APPENDIX.
it without a remainder ; then the cube of the quotient, multiplied by the cube
of the number found in the table, will give the answer.
Example. — What is the cube of 2700? 2700, being divided by 900, the quo-
tient is 3, the cube of which is 27 and the cube of 900 is 729,000,000, there-
fore :
27 x 729,000,000 = 19,683,000,000 : the Ans.
To find the square or cztbe root of numbers higher than is found in the table.
Rule. — Select, in the column of squares or cubes, as the case may require, that
number which is nearest the given number ; then the answer, when decimals
are not of importance, will be found directly opposite, in the column of num-
bers.
Example. — What is the square root of 87,620? In the column of squares,
87,616 is nearest to the given number ; therefore, 296, immediately opposite
in the column of numbers, is the answer, nearly.
Another example. — What is the cube root of 110,591? In the column of
cubes, 110,592 is found to be nearest to the given number ; therefore, 48, the
number opposite, is the answer, nearly.
To find the cube root more accurately. Rule. — Select from the column of
cubes that number which is nearest the given number, and add twice the
number so selected to the given number ; also, add twice the given number
to the number selected from the table. Then, as the former product is to the
latter, so is the root of the number selected to the root of the number given.
Example. — What is the cube root of 9200? The nearest number in the col-
umn of cubes is 9261, the root of which is 21, therefore :
9261 9200
2 2
18522 18400
9200 9261
As 27,722 is to '27,661, so is 21 to 20-953 + , the Ans.
Thus, 27,661 x 21 = 580,881, and this divided by 27,722 = 20-953 -f .
To find the square or cube root of a whole number with decimals. Rule. — Sub-
tract the root of the whole number from the root of the next higher number,
and multiply the remainder by the given decimal ; then the product, added to
the root of the given whole number, will give the answer correctly to three
places of decimals in the square root, and to seven in the cube root.
Example. — What is the square root of 11-14? The square root of n is
3-3166, and the square root of the next higher number, 12, is 3-4641 ; the for-
mer from the latter, the remainder is 0-1475, and this by 0-14 equals 0-02065.
This added to 3-3166, the sum, 3-33725, is the square root of 11-14.
To find the roots of decimals by the use of the table. Rule. — Seek for the
given decimal in the column of numbers, and opposite in the columns of roots
will be found the answer, correct as to the figures, but requiring the decimal
point to be shifted. The transposition of the decimal point is to be performed
thus : For every place the decimal point is removed in the root, remove it in
the number t^vo places for the square root and three places for the cube root.
THE REDUCTION OF DECIMALS. 647
Examples. — By the table, the square root of 86-0 is 9-2736, consequently by
the rule the square root of 0-86 is 0-92736. The square root of 9- is 3-, hence
the square root of 0-09 is 0-3. For the square root of 0-0657 we have
0*25632, found opposite No. 657. So, also, the square root of 0-000927 is
0-030446, found opposite No. 927. And the square root of 8-73 (whole num-
ber with decimals) is 2-9546, found opposite No 873. The cube root of 0-8
is 0-928, found at No. 800 ; the cube root of 0-08 is 0-4308, found opposite
No. 80, and the cube root of 0-008 is 0-2, as 2-0 is the cube root of 8-0. So
also the cube root of 0-047 's 0-36088, found opposite No. 47.
RULES FOR THE REDUCTION OF DECIMALS.
«•,
To reduce a fraction to its equivalent decimal. Rule. — Divide the numerator
by the denominator, annexing cyphers as required.
Example. — What is the decimal of a foot equivalent to three inches ?
3 inches is j\ of a foot, therefore :
•& . . . 12)3-00
•25 Ans.
Another example. — What is the equivalent decimal of $ of an inch?
1 ... 8)7-000
•875 Ans.
To reduce a compound fraction to its equivalent decimal. Rule. — In accordance
with the preceding rule, reduce each fraction, commencing at the lowest, to
the decimal of the next higher denomination, to which add the numerator of
the next higher fraction, and reduce the sum to the decimal of the next higher
denomination, and so proceed to the last ; and the final product will be the
answer.
Example. — What is the decimal of a foot equivalent to five inches, f and TV
of an inch ?
The fractions in this case are, i of an eighth, | of an inch, and Trv of a foot,
therefore :
eighths.
inches.
•rV 12)5-437500
-453125 Ans.
The process may be condensed, thus : write the numerators of the given
648 APPENDIX.
fractions, from the least to the greatest, under each other, and place each de«
nominator to the left of its numerator, thus :
8
3. CQOO
12
5-437500
•453125 Ans.
To reduce a decimal to its equivalent in terms of lower denominations. Rule.
— Multiply the given decimal by the number of parts in the next less denomi-
nation, and point off from the product as many figures to the right hand as
there are in the given decimal ; then multiply the figures pointed off by the
number of parts in the next lower denomination, and point off as before, and
so proceed to the end ; then the several figures pointed off to the left will be
the answer.
•Example, — What is the expression in inches of 0-390625 feet?
Feet 0-390625
12 inches in a foot.
Inches 4-687500
8 eighths in an inch.
Eighths 5 • 5000
2 sixteenths in an eighth.
Sixteenth i-o
Ans., 4 inches, £ and •£$.
Another example. — What is the expression, in fractions of an inch of
0-6875 inches?
Inches 0-6875
8 eignths in an inch.
Eighths 5-5000
2 sixteenths in an eighth.
Sixteenth i-o
Ans., f and ^
TABLE OF CIRCLES.
(From Gregory's Mathematics.)
FROM this table may be found by inspection the area or circumference of a
circle of any diameter, and the side of a square equal to the area of any given
circle from i to 100 inches, feet, yards, miles, etc. If the given diameter is in
inches, the area, circumference, etc., set opposite, will be inches ; if in feet,
then feet, etc.
Diam.
Area.
Circum.
Side of
equal sq.
Diam.
Area.
Circum.
Side of
equal sq.
•25
•04908
•78539
•22155
•75
90-7625?
33-7721-2
9-52693
•5
•19635
1-57079
•44311
11-
95-03317
34-55751
9-74S1'.*
•75| '44178
235619
•66467
•25
99-40195
35-34291
9-97005
]•
•78539
3-14159
•88622
•5
103-86890
36-12331
10-19160
•25
1-22718
3-92699
1-10778
•75
108-43403
36-91371
10-41316
•5
1-76714
4-71238
1-32934
12-
11309733
37-69911
10-63472
•75
2-40528
5-49778
1-55089
•25
117-85831
3848451
10-85627
2-
3-14159
6-28318
1-77245
•5
122-71846
3926990
11-07733
•25
3-9760?
7-06858
1-99401
•75
127-676281 40-05530
11-29939
•5
4-90873
7-85393
2-21556
13-
132-73228
40-84070
11-5-2095
•75
5-93957
8-63937
2-43712
•25
137-88646
41-62610
11-7425;)
3-
7-068581 942477
2-65868
-.-)
14313881
42-41150
11-96406
•25
8-29576
10-21017
2-88023
•75
148-48934
43-19689
12-1856-2
•5
9-62112
10-99557
3-10179
14-
153-93804
43-98229
12-40717
•75
11-044661 11-78097
3-32335
•25
159-48491
44-76769
12-62S73
4-
12-56637 12-56637 3-54490
•5
165-12996
45-55309
12-85029
•25
14-186251 13-35176J 3-76646!
•75
170-87318
46-33849J 13-07184
•5
15-90431
14-13716| 3-98802
15-
176-71458
47-12338! 13-21)340
•75
17-72054
14-922561 4-20957!
•25
182-65416
47-909281 13-51496
5-
19-63195
15-70796 4-431 13j
-.r>
188-69190
48-69468 13-73651
•25
21-64753
16-49336 4-65269J
•75
194-82783
49-48003 13-95307
•5
23-75829
17-27875
4-87424
IS-
201-06192
50-26548 14-17963
•75
25-96722
18-06415
5 -095 WO,
•25
207-:W4'20
51-05088: 14-101 IS
6-
28-27433
18-84955 5-31736
•5
213-82464
51-83027 14-62274
•25
30-67961
19-63495! 5-53891
•75
220-35327
5262167
14-84430
•5
33-18307
20-42035 5-76047 17-
226-98006
53-40707
15-065-»5
•75
35-78470
21-20575. 5-9S2()3i' -25
233-70504
54-19247
15-28741
7-
33-48456
21-99114
6-20358! ! -5
240-52818
54-97787 15-50897
•25
41-28249
27-77654
6-42514 j -75
•217-44950
55763261 15-73052
.K
44-17H61
23-56194
6-646701: 18-
254-46900
56-54866! 15-9.V20S
•75
47-17297
24-34734
6-86825
•25
261-58667
57-33406
16-17361
8-
50-26518
25-13274
7-08981
•5
26880252
53-11946
1639519
25
53-45616
25-91813
7-31137
75 27()-llC,54
58-90486
16-61675
•5
5S-74501
26-70353
7-53292
19- 2-<3-52873
59-69026
16-8:KH
•75
60-13204
27-48893 7754 18
•25 291-03910
6047565
17-05986
9-
63617-2")
28-27433 7-976J4
•5 i 298-64765
61-26105
17-28 U2
•25
07-20063
29-0.'>97* 8-197.')9
•75
300-:{513?
62-04645
17-5;»-2'J8
•5
70-83218
29-84513 8-41915
20-
314-15926
62-83185
17-72453
•75
74-66191
30-63052 8-64071
"J.-> 322-00-233
6361725
17-94609
10-
78-53981
31-41592 886226
•5
330-06357
64-40264
18-16765
•25
82-51589
38*01321 9-08382
•75
338-16-299
65-18804
18-38920
•5 86-59014
32-986721 9-3i):>33 21- I 346-36059
6 j 97341
18-61076
650
APPENDIX.
)iam.
Area.
Circam.
Side of
equal sq.
Diam.
Area.
Side of
Circum. enual sq.
21-25
354-65635
66-75834
1883232
38-
1134-114941 119-38052! 3367662
5
363-05030
67-54424
19-05337
•25
1149-086601 120- 16591 i 33 893 17
•75
371-54241
68-32964
19-27543
•5
1164-15642
120-95131 34- IS 973
22- 380-13271
69-11503
19-49699
•75
1179-32442 121-73671 3434129
•25 1 388-82117
69-90043
19-71854
39-
1194-59060 122-52211
34-56285
•5
337-60782! 70-68583
19-94010
•25
1209-95495
123-30751
34-78440
•75
406-49263
71-47123
20-16166
•5
1225-41748 124-09290
35-00596
23-
415-475^2
72-25663
20-38321
•75
1210-97818
124-878301 35-22752
•25
424-55679
73-04202
20-60477
40-
1255-63704! 125-663701 35 44907
•5
433-736131 73-82742
20-82633
•25
1272-39411
126-449101 35-67063T
•75
443-01365| 74-61282
21-04788
•5
1288-24933
127-23450 35-89219
24-
452-38934J 75-39822
21-26944
•75
: 1304-20273
128-01990 36-11374
•25
461-86320
76-18362
21-49100
41-
1320-25431 123-80529J 36-33530
•5
471-43524
76-96902
21-71205
•25
1336-40406
129-59069 36-55686
•75
481-105461 77-75441
2193411
•5
1352-65198
1 30-3760') 36-77841
25-
490-87385! 78-53981
22-15567
•75
1368-99808
131-16149 36-99997
•25
500-740411 79-32521
22-37722
42-
1385-44236
131-94689 37-2*2153
•5
. 510-70515! 80-11061
22-59878
•25
1401-98480
13273228 37-44308
•75
520-76306! 80-89601
22-82034
•5
1418-625431 133-51768! 3766464
26-
530-92915 81-68140
23-04190
•75
1435-36423
134-303081 37-88620
•25
541-18842
82-46680
23-26345
4o-
1452-20120
135-08348 38-10775
•5
551-54586
83-25220
23-48501:
•25
1469-13635
135-87338 38-32931
•75
562-00147
84-03760
2370657
•5
1486-16967
136-65928! 38-55087'
27-
572-55526
84-82300
23-92812;
•75
1503-30117
137-44467J 33-77242
•25
583-20722
85-60839
24-14968
44-
1520-53084
138-83007 38-993^8
•5
593-95736
86-39379
24-37124
•25
1537-85869
1P9-01547J 39-21554
•75
604-80567
87-17919
24-59279;
•5
1556-23471 139-80087! 39-43709
28-
615-75216
87-96459
24-81435!
•75
1572-80890
140-58627
3965865
•25
626-79682
88-74999
25-03591
45-
152043128
141-37166
39-88021
•5 637-93965
89-53539
25-25746
•25
1608-15182
142-15706 40-10176
•75
649-18066
90-32078
25-47'J02
•5
1625-97054
142-94246! 40-32332
29-
66051985
9M0613
25-70058
•75
164388744
143-727861 4054488
•25
671-95721
9 39153
25-92213
46-
1661-90251 14451326! 40-76643
•5
683-49275
92-67698
26-14359
•25 1680-01575 145'29866l 40-98799
•75
695-12646
93-46233
26-36525
•5 1698-22717
146-08405 41-20955
30-
706-85834
94-24777
26-58680
•75 1716-53677
146-86945 41-43110
•25
71868840
9503317
26-80836
47- 1 1734-94454
147-65485 41-65266
•5
730-61664
95-81857
27-02992
•251 1753-45048
148-44025! 4187422
•75
74264305
96-603971 27-25147;
•5
1772-05460
149-22565
42-09577
31-
751-76763 97-38937; 27-47303 i -75 1790-75639
150-01104
4231733
•25
766-99039" 98-17477
27-69459! 48- i 180955736
150-79644
42-53889
•5
779-31132
9896016
27-91614; -25 '' 1828-45601
151-58184
42-76044
•75
791-73043
99-74556
28-13770
•5 1847-45282
152-36724
42-98200
32-
804-24771
100-53096
2S-M3926; -75 1866-54782 153-15264
43-20356
•25
816-86317
101-31636
2M/59081I 49-' 1885-74099 153'93804
4342511
•5
829-57631
102-10176
28-h0237; -25! 1905-83233 154-72343
43-64667
•75
842-38861
102-887151 2902393
•5 1924-42184
155-50883 43-86823
*33-
855-29859
10367255
29-24548
•75
1943-90954
156-29423 44-08978
•25
868-30675
104-45795
29-46704!
50-
1963-49540
157-07963] 44-31134
•5
881-41308
105-24335
29-68860
•25
1983-17944
157-96503] 44-53290
•75
89461759
106-028751 29-91015
•5
2002-96166
153-65042 44-75445
34-
907-92027 106-81415
30-13171
•75! 2022-84205
159-43582! 44-97601
25
921-321131 107-59954
30-35327
51-
2042-82062
160-22122
45-19757
•5
934-82016
308-33494
30-57482
•25
2062-89736
161-00662
45-41912
•75
94841736
109-17034
30-796'*8
•5 ! 2083-07227
161-79202
45-64068
35-
962-11275
109-95574
31-01794J ! -751 210334536
162-57741
43-85224
•25
975-90630
110-74114
31-23949 52- 1 2123-71663
163-36281
46-08380
•5
989-79803
111-52653! 31-46105
•25| 2144-18607
164-14821
46-30535
•75
1003-78794
112311931 31-68261
•5 1 2164-75368
184-93361
46 52691
36
1017-87601
11309733
31-90416
•75! 2185-41947
165-71901
46-74847
•25
1032-06227
113-88-273
32-12572
53- 2206-18344
166-50441
46-97002
•5
•75
1046-34670 11466813
1060-729301 115-45353
32-34728
32-56383
•25 2227-04557
•5 2248-00589
167-2-8980
168-07520
47-19158
47-41314
37-
1075-21008
116-23892
32-7903J
•75
2269-06433
168-86060
47-63469
•25
1089-78903
117-024 3-2
33-01195
54-
2290-22104
169-64600
47-85625
•5
1104-46616
117-80972
33-23350
•25: 2311-475^8
170-43140
48-07781
•75
1119-24141- 118-59572
33-45506
•5 ! 2332-82889 171-21679 48-29936
TABLE OF CIRCLES.
65I
Side of
&<leo/ '
Du.ni.
Arciu
Circuir,,
equal sq. Diam. Area.
Circum.
equal ••!.
"sflb
2354-2800S 172-00219
"~4~8l>2092
71-5
4015-1517f>
~1^^23^! ""63-36522
55-
2375-82344 172-78759
48-74248
•75
404327833
225-40927 i 63-58678
•25
2397-47698! 173-57299
48-96403
72-
4071-50407
226-19467
63-80333
•5 2419-22269 174 -3583 J
49-18559
•25
4099-82750
226-98006
64-02989
•75 2141-0665?! 175-14379
49-40715
•5
4128-24909
227-76546
64-25145
56-
246300864! 175-92918
49-62870
•75
4156-76886
22855086
64-47300
•25
2185-04887' 17<r7145Sl 49-85026
73-
4185-38681
229-33626
64-69456
•5
2507-18728 177-49998 5007182
•25
4214-10293
230-12166
64-91612
•75 2520-452337 i 178-28538! 50-29337
•5
4242-91722
230-90706
65-13767
57- 2551-75363I 179-07078 50-51493
•75
4271-82969
231-69245
65-35923
•25| 2574-19156 179-85617
50-73649
74-
4300-84034
232-47785
65-58079
•5
2596-72267 180-64157
. 50-95804
•25
4329-94916
23326325
65-80234
•75
2619-35196 181-426b7| 51-17960
•5
4359-15615
234-04865
66-02390
58-
264207942! 18221237 51-40116
•75
4388-46132
23483405
66-24510
•25| 26(54 90505; 182-99777
51-62271
75-
4417-86466
235-61944
66-46701
•5
2687-82836 183-78317
51 84427
•25
4447-36618
236-40484
66-6885?
•75
2710-85084 - 184-56856
52-06583
•5
4476-965S8
237-19024
66-91043
59-
2733-97100 185-353 J6
52-28738
•75
4506-66374
237-97f)64
67-13168
•25
275718933 18613936
52-50894
76-
4536-45979
238-76104
67-3532 1
•5
2780-50584 186-92476
52-73D50
•25
456635400
239-54643
67-57480
•75
280392053 187-71016
52-95205
•5
45D6 34640
240-33183
67-79635
60-
2327-43338 188-49555 53-17364
•75
462643696
241-11723
6301791
•25
2851-04442 189-28095
5339517
77-
465562571
241-90263
68-23J47
•5
2874-75362 193-06635
5361672!
•25
468691262
242-6H8U3
68-46102
•75
2898-56101) 193-85175
53-83328 !
•5
471729771
243-47343
68-68258
61
2922-46656
191-63715
54-05H84!
"75
4747-78098
24425382
68-90414
•25
2946-47029
192-42255
5428139
78-
477836242
245 04422
63-12570
•5
2970-57220
193-20794
5451)295
•25
4809-04204
24582962
6931725
•75
2994-77228
193-99334
54-724511
•5
4839-81983
24661502 69-56881
62-
3019-07054
194-77874
54<MG','6
•75
4870-79579
247-40042 69-79037
•25
304346697
195-56414
55-167621
79-
4901-66993
248-18591
70-01192
•5
3J67-(J6I57
196-34954
55-33918
•25
4;>32-74225
248-97121
70-23318
•75
309255135
197-13493
5561073
•5
4963-91274
24975661
70-45504
63
3117-21531
197-92033 55-83229
•75
4995-18140
250-34201
70-67659
•25
314203444
198-70573 55-053-5
80-
5026-54824
251-32741
70-39815
•5
3166-92174
199-49113 56-27510
•25
505801325
252-11281
71-11971
•75
3191-90722
200-27653
56-49698
•5
503957644:
252-89820
71-34126
64-
3*16-99087
201-061^2
5671852
•75
5 121 23781
25368360
71-56282
•25
3242-17270
201-84732 56-9100?
81-
5152-99735
254*46900
71-78433
•5
326745270
20263272 57-16163
•25
5184-85506
255-25440
72-00593
•~5
3292 83088
203-41812 57-3 S3 19
•5
5216-SI<t'.l.~>
!2^6-()3JSO
72 -22749
;35-
3318-30784
204-20352 57-60475
•75
5248-86501;
25682579
72-14905
•25
334388176
204-98892 57-8263<)|
82-
5281-01725
257-61059
• 2-67060
•5
3369-55447
20577431
5804786'
•25
531326766
258-3i)599
72-89216
•75
339532534
206-55971
53-26942
•5
5345-r.ir.-ji
259-18139
7311372
66-
3421 19439
207-34511
5349097
•75
5378-06301
259-96679
?:<:«52?
25
344716162
203-130511 53-712:-):!
83-
541060794
260-75219
? 3 55633
•5
3473 22702
208-915911 58-93 lO'.i
•25
:»m-2f>K!:i
261-53753
73 77839
•75
349939060
209-7013!) 5.H:V>!il
•5
r.l?.Vii9234
262-32298
73*99994
67-
3525-65235
210 48670
59-37720
•75
5508-83180
263-10338
74-22150
25
3552-01228
211-27210
595;»S76
84-
5.') 1 1 ?tV.t 1 1
263-89378
71:14306
•5
3573-47033
212-05750
59-821)31
•25
55?480f>2:>
264-67918
74-66461
*T5
3505 026i55
212-84290 60-04 I*?
•5
5607-US923
265-46457
74-8861?
68-
3631-68110
2i3-r,->>3o r,o-2i;:m
•75
5641-17I3J
266-24997
75-10773
•25
9658-43373
214-4 K«iy CO-l-il'.tS
85-
."•67 1-50 173
267-03537
75'32:'->
•5
S68528453
21.")- 19.109 60-70C,;:!
•25
5707-y;{o-j:j
2G7-82077
75-55: iM
•75
371!} 23350
2J598449 60-92SIU
•5
5741 1 :,r,. -j
268-60617
75-772 li »
69-
3^39-28065
216761IS9 C.l-ll.G.")
•75
5775-08178
269-3915?
75-99395
•25
'576 C -42597
217-f)5;-.29 61-37121
86-
:.s(is-H0481.
270-176% 7621551
•5
379JJ-66947
21834068
61-59277
•25
5Sl262(i<)2
270-96236 764370?
•75
332 '01 115
21.
61-81132
•5
5876-54540
271-74770
76-65362
70
:MS-|.-,KM) vMH-'jiiis1 62-03588
•75
5.nor,r,29r.
272-53316
76-88018
25
3S7:>-<J3902j 2-20-tVjr.8S 62"J:>71I
87-
:.'J 1 1 67869
27331856
77- 10174
•5
3i)03-62522 2 i 1-43228 62-478'J9
•25
5978-892C.O
274-10395
77-:«:*2'.»
•75
3931-35959 MB-JJCTW 62-70055
•5
6013-20468
274-88935
77*54485
71
3959-19214' •J-.'^or.WT r>2'J2211
•75
604761 J91
27567475
7776641
•2?
3987- 12286 223 &i i47 03 1 136l>
88-
6082-12337,
276-16015
77-98796
652
APPENDIX.
Diam.
Area.
Circum.
Side of
equal sq.
Dinm.
Area.
Circum.
Side ot
equal sq.
88-25
6116-729931 277-24555
78-20952
94-25
6976-74097
2-6-09510
83-52688
•5
6151-43476 27803094
78-43103
•5
7012-80194
296-88050
83-74844
•75
6186-23772 278-81634
7865263
•75
7050-96109
297-66590
83-97000
89-
6-22M3385 279-60174
78-87419
95-
7083-21842
298-45130
84-19155
•35! 6256-13S15I 23033714
79-09575
•25
7125-57992
299-23070
84-4131 1
•5
6291-23563
231-17254
79-31730
•5
7163-02759
300-02209
84-63467
•75
63-26-43129
281-95794
79-53386
•75
7200-57944
300-80749
84-85622
90-
6361-7-2512
28-2-74333
79-76042
96-
7233-22947
3U 1-59239
85-07778
•25
6397-11712
233-52873
79-98198
•25
7275-97767
302-37829
85-291)34
•5
6432-60730
284-31413
80-20353
•5
7313-82404
303-16369 85-52039
•75
6463-19586
285-09953
80-42509
•75) 7351-76859
303-94908
85-74245
91-
6503-83219
285-83493
80-64669
97-
7389-81131
304-73448
85-96401
.25
653J-66639
286-67032
80-85820
•25
7427-95221
305-51988
86-18556
•5
6575-54977
237-45572
81-08976
•5
7466-19129
306-30523
86-40712
•75
6611-53382
288-24112
81-31132
•75
7504-52853
307-09068
86-62868
92-
6647-61005
239-02652
81-53287
98-
7542-96396
307-87608
86-85023
•25
6683-78745
289-31192
81-75443
•25
7581-49755
- 308-66147
87-0717t'
•5
6720-06303
2:)0-51J7:)'2
81-975^'J
•5
7620-12933
309-44637
87-29335
•75
6756-43678
291-33271
82-VJ754
•75
7653-85927
310-23227
87-51490
93-
6792-90871
292-16811
82-41910
99-
7697-68739
311-01767
87-73646
•25
6829-47831
292-95351
82-04066
•25
7736-61369
311-80307
87-95802
•5
6866-14709
293-7339}
82-86-221
•5
7775-63816
312-58346
88-17957
•75
6902-91354
294-52431
83-08377
•75
7814-760311 313-37336
88-40113
94-
6939-77317 295-30970, 83 30533 jj 100-
7653-98163| 314-15926
88-62269
The following rules are for extending the use of the above table.
To find the area, circumference, or side of equal square, of a circle having a
diameter of more than 100 inches, feet, etc. Rule. — Divide the given diameter by
a number that will give a quotient equal to some one of the diameters in the
table ; then the circumference or side of equal square, opposite that diameter,
multiplied by that divisor, or the area opposite that diameter, multiplied by
the square of the aforesaid divisor, will give the answer.
Example. — What is the circumference of a circle whose diameter is 228
feet ? 228, divided by 3, gives 76, a diameter of the table, the circumference
of which is 238-761, therefore :
238-761
3
716 -283 feet. Ans.
Another example. — What is the area of a circle having a diameter of 150
inches? 150, divided by 10, gives 15, one of the diameters in the table, the
area of which is 176-71458, therefore:
176-71458
100 = TO X IO
17,671-45800 inches. Ans.
To find the area, circumference, or side of equal square, of a circle having an.
intermediate diameter to those in the table. Rule. — Multiply the given diameter
by a number that will give a product equal to some one of the diameters in
the table ; then the circumference or side of equal square opposite that diame-
ter, divided by that multiplier, or the area opposite that diameter divided by
the square of the aforesaid multiplier, will give the answer.
CAPACITY OF WELLS, CISTERNS, ETC. 653
Example. — What is the circumference of a circle whose diameter is 6J, or
6-125 inches? 6-125, multiplied by 2, gives 12-25, one of the diameters of the
table, whose circumference is 38-484, therefore :
2)38-484
19-242 inches. Ans.
Another example. — Wh?t is the area of a circle, the diameter of which is 3-2
feet? 3-2, multiplied by 5, gives 16, and the area of 16 is 201-0619, therefore :
5 x 5 = 25)201-0619(8-0424 + feet. Ans.
200
106
100
61
50
Note. — The diameter of a circle, multiplied by 3-14159, will give its cir-
cumference ; the square of the diameter, multiplied by -78539, will give its
area; and the diameter, multiplied by -88622, will give the side of a square
equal to the area of the circle.
TABLE SHOWING THE CAPACITY OF WELLS, CISTERNS, ETC.
The gallon of the State of New York, by an act passed April n, 1851, is required to conform to
the standard gallon of the United States government. This standard gallon contains 231 cubic
inches. In conformity with this standard the following table has been computed.
One foot in depth of a cistern of
3 feet diameter will contain ........................ 52 • 872 gallons.
3* " " ....................... 7I-965
4 " ........................ 93-995 "
4i " ........................ 118-963
5 " " ........................ 146-868
5i " " ........................ 177-710
6 " " ..................... 211-490 "
6£ " " ....................... 2*8-207
7 " " ....................... 287-861 "
8 " " ....................... 375-982
9 ....................... 475-852
10 " " ............ : ............ 587-472
12 " " ....................... 845-959
Note. — To reduce cubic feet to gallons, multiply by 7-48. The weight of a
gallon of water is 8-355 Ibs. To find the contents of a round cistern, multi-
ply the square of the diameter by the height, both in feet, and this product by
5-875-
654
APPENDIX.
TABLE OF WEIGHTS.
MATERIALS USED IN THE CONSTRUCTION OR LOADING OF
BUILDINGS.
WEIGHTS PER CUBIC FOOT.
As per Barloiv, Gallier, Jlaswell, Jfurst, Rankine, Tredgold, Wood
and tJie AutJwr.
MATERIAL.
|
In
£
AVERAGE.
MATERIAL.
S
0
K
to
o
H
AVERAGE.
WOODS.
Mahogany, St. Domingo. . . .
45
65
49
55
41
41
51
46
Mulberry
35
55
45
Alder
35
51
38
Oak, Adriatic
62
49
51
5O
" Black Bog
60
66
63
Ash
41
57
49
" Canadian
54
Beech
46
47
Birch.
35
49
42
" English ..
S
7°
54
Box .
59
65
62
Live
57
79
68
83
" Red
47
51
64
" White.
50
Cedar
27
35
31
Olive..
58
47
52
44
" Palestine
u Virginia Red ......
Cherry
30
32
. 38
46
34
40
39
Pear-tree . .'
Pine, Georgia (pitch)
'* Mar Forest
40
38
\
42
48
43
Chestnut, Horse
29
35
" Memel and Riga
29
**
32
Sweet
Cork . ...
27
55
41
15
" Red
" Scotch
27
37
39
Cypress .
27
34
" White
28
*' Spanish
Deal, Christiania. .
40
44
" Yellow
27
41
39
33
45
'* English
41 (Norway Spruce).
21
t9
7
Poplar
23
37
30
44
Dogwood
47
23
Ebony
69
8^
76
45
Elder
43
30
Elm
31
59
46
Satinwood
e;5
57
27
oA
30
" (Red Pine)
3°
37
36
38
" Riga
47
Teak
61
51
53
Tulip-tree
30
" ' Water
62
Vine.
81
so
37
Walnut Black
• 26
33
Hemlock
21
26
" White'.
eg
49
40
la
49
Whitewood
27
te
52
Yew
50
Larch . ...
31
33
" Red
31
43
METALS.
" White
23
Bismuth, Cast
614
41
83
62
487
5O6
41
-?6
544
57
k< Plate
q28
531
35
38
Bronze .
508
516
WEIGHT OF MATERIALS.
655
TABLE OF WEIGHTS.— (Continued.)
MATERIALS USED IN THE CONSTRUCTION OR LOADING OF
BUILDINGS.
WEIGHTS PER CUBIC FOOT.
As per Barlow, Gallier, Jfaswell, Hurst, Rankine, Tredgold, Wood
and the Author.
MATERIAL.
Z
1
o
H
AVERAGE.
MATERIAL.
I
t*
O
H
AVERAGE.
Copper Cast
537
549
543
06
u_
1O4
556
dry
1OO
" Plate
644
44 in Cenfent .
112
Gold
1206
" in Mortar
1 1O
110S
Caen Stone
13O
509
81
475
487
481
" Roman, Cast.
1001
Cast
434
474
454
" Malleable
" Wrought
486
475
480
equal parts..
Chalk
116
113
14 *
1 Lead Cast
709
Clay
122
Er^lish Cast
717
" with Gravel
16O
" M ficd
713
90
851
?6
102
83
" " 60°
849
8i
79
" " 212°
837
85
Nickel Cast
488
Coke
^.6
62
54 I
453
Concrete, Cement
13O
975
1O6
1345
U o
n •' Rolled
1379
126
142
" with Gravel
126
636
25O
*' ' Pure Cast
655
Feldspar
16O
" " " Hammered
" Standard
...
658
644
Flagging, Silver Gray. . . .
Flint
- -
185
163
Steel
<86
402
489
l6e
160
Tin Cast
!,6
1 •<
462
" Flint
183
429
449
439
" Green
l«i;»
Zinc, Cast
'« Plate
163
" White
167
181
174
STONES, EARTHS, ETC.
Granite
nR
Hi 5
161
x6c
180
173
166
156
" Guernsey
185
8O
166
277
Gnvel
9°
12O
105
Basalt
187
134
J!-5
i •{ M
135
1^5
14O
t)atn i one..^..
129
Lime Unslaked
52
r>f c r'
16O
11Q
169
Brick '
85
102
Aubigne
146
138
162
,i -»T TJ h ' °d
107
Marble
161
178
17O
" " Salmon
'
1OO
166
105
17O
656
APPENDIX.
TABLE OF WEIGHTS.— (Continued^
MATERIALS USED IN THE CONSTRUCTION OR LOADING OF
BUILDINGS.
WEIGHTS PER CUBIC FOOT.
As per BarloTV, Gallier, Haswell, Hurst, Rankine> Tredgold, Wood
and the Author.
MATERIAL.
1
S
AVERAGE.
MATERIAL.
6
N
CK
o
H
AVERAGE.
Marble Enstchester
167
i63
IOO
110
178
169
179
140
173
167
166
167
140
125
175
155
98
103
107
9
86
105
1OO
118
83
146
72
8O
180
175
147
56
165
165
124
112
1O5
105
123
1O5
97
172
126
144
133
142
134
141
134
162
150
Serpentine
" Chester Pa
...
165
144
152
95
159
167
157
18O
135
151
14O
16O
14O
124
115
170
76
SS
49
59
62
26
2O
58
57
61
17
61
69
114
73
131
559
68
171
133
131
14
8O
62}
64
81
Egyptian
" French
Shingle
137
181
Marl
Slate
Mica
" Cornwall
109
118
" Welsh
87
83
Stone, Artificial
120
150
" Paving . ,.„
Stone-work .
120
i€o
" Hair, incl. Lath and
Nails, per foot sup.
7
ii
" Rubble
Sulphur, Melted
" Sand 3 and Lime paste 2
" " 3 " " " 2
well beat together. .
Peat Hard
Trap Rock
MISCELLANEOUS.
Petrified Wood
Pitch
Plaster Cast
Bark, Peruvian
" ' Red
132
161
17
»4
34
25
62
24
66
Charcoal
Pumice-stone
Cotton, baled. . .
Fat
52
JO
56
Rotten-stone
Sand, Coarse
92
90
118
ii8
120
128
Gutta-percha
Hay baled
India Rubber
' Dry
Moist
Pit
92
101
Quartz
Red Lead
158
130
Rock Crystal
" Amherst, O. . . .
" Belleville, N. J...
Berea, O
" Dorchester, N. S
" •« Little Falls, N. J.
u Marietta, O. . . .
" Middletown, Ct..
Salt
20
100
Saltpetre —
Snow
"a
60
Water, Ram
" Sea
Whalebone
INDEX.
Abscissas of Axes, Ellipse 484
Abutments, Bridges,. Strength of. 227
Abutments, Houses, Strength of.. 53
Acute Angle Denned 349, 544
Acute-angled Triangle Denned... 545
Acute or Lancet Arch 51
Algebra, Addition 398
Algebra, Application of 393
Algebra, Binomial, Multiplica-
tion of , 409
Algebra, Binomial, Square of a... 429
Algebra, Binomial, Squaring a. . . 410
Algebra Defined 392
Algebra, Denominator, Least
Common 404
Algebra, Division, the Quotient.. 419
Algebra, Division, Reduction.... 419
Algebra, Division, Reverse of
Multiplication 418
Algebra, Factors, Multiplication
of Two and Three 409
Algebra, Factors, Multiplication
of Three . 408
Algebra, Factors, Squaring Differ-
ence of Two 412
Algebra, Fractions Added and
Subtracted 403
Algebra, Fractions, Denominators 407
Algebra, Fractions Subtracted... 405
Algebra, Hypothenuse, Equality
of Squares on 416
Algebra, Letters, Customary Uses
of 396
Algebra, Logarithms Explained.. 425
Algebra, Logarithms, Examples in 426
Algebra, Multiplication, Graphical 408
Algebra, Progression, Arithmeti-
cal... 432
Algebra, Progression, Geometrical 435
Algebra, Proportion Essential 347
Algebra, Proportionals, Lever
Formula 421
Algebra, Quantities, Addition and
Subtraction 424
Algebra, Quantities, Division of. 424
Algebra, Quantities, Multiplica-
tion of 424
Algebra, Quantities with Negative
Exponents 423
Algebra, Quantity, Raising to any
Power 423
Algebra, Radicals, Extraction of.. 425
Algebra, Rules are General 394
Algebra, Rules, Useful Construc-
ting 394
Algebra, Signs 397
Algebra, Signs, Arithmetical Pro-
cess by 396
Algebra, 'Signs, Changed when
Subtracted .... 400
Algebra, Signs, Multiplication of
Plus and Minus. .. 415
Algebra, Squares on Right- Angled
Triangle 417
Algebra, Subtraction 398
Algebra, Sum and Difference, Pro-
duct of 413
Algebra, Symbols Chosen at Pleas-
ure 395
Algebra, Symbol, Transferring a.. 399
Algebra, Triangle, Squares on
Right-angled 417
658
INDEX.
Alhambra, or Red House, Ancient
Palace of the n
Ancient Cities, Historical Ac-
counts of 6
Ancient Monuments, their Archi-
tects 6
Angle at Circumference of Circle. 358
Angle Defined 544
Angle to Bracket of Cornice, To
Obtain 343
Angle, To Measure a, Geometry.. 348
Angle rib to Polygonal Dome. . . . 223
Angle-rib, Shape of Polygonal
Domes 223
Amulet or Fillet, Classic Mould-
ing 323
Antae Cap, Modern Moulding. ... 334
Antique Columns, Forms of 48
Antiquity of Building 5
Arabian and Moorish Styles, An-
tiquities of ii
Araeostyle, Intercolumniations. . . 20
Arc of Circle Defined 547
Arc of Circle, Length, Rule for. . 475
Arc, Radius of, To Find 561
Arc, Versed Sine, To Find (Geom-
etry) 561
Arcade 52
Arcade of Arches, Resistance in. . 52
Arcade in Bridges, Strength of
Piers 52
Arch 50
Arch, Acute or Lancet 51
Arch, Archivolt in 52
Arch, Bridge, Pressure on 51
Arch, Building, Manner of 50
Arch, Catenary 51
Arch, Construction of 50
Arch, Definitions and Principles of 52
Arch, Extrados of 52
Arch, Form of 50
Arch, Formation in Bridges 51
Arch, Hooke's Theory of an 50
Arch, Horseshoe or Moorish 51
Arch, Impost in 52
Arch, Intrados of 52
Arch, Keystone, Position of 50
Arch, Lateral Thrust in . 52
Arch, Ogee 51
Arch, Rampant 51
Arch, Span of an 52
Arch, Spring in an 52
Arch, Stone Bridges 230
Arch-stones, Bridges, Jointing. .. 233
Arch, Strength of 50
Arch of Titus, Composite Order. . 28
Arch, Uses of 50
Arch, Voussoir in 52
Architect and Builder, Construc-
tion Necessary to 56
Architect, Derivation of the Word 5
Architects of Italy, I4th Century. 12
Architecture, Classic Mouldings
in 323
Architecture, Ecclesiastical, Origin
of 14
Architecture, Egyptian, Character
of 33
Architecture, Egyptian, Features
of 30
Architecture, English, Ccttage
Style 35
Architecture, English, Early n
Architecture, Grecian and Roman 8
Architecture, Grecian, History of. 6
Architecture, Hindoo, Character
of . 30
Architecture, Order, Three Princi-
pal parts of 14
Architecture, Principles of 44
Architecture, Roman, Ruins of... II
Architecture in Rome Defined. .. 7
Architecture, Result of Necessity. 13
Architrave Defined 15
Area of Circle, To Find 475
Area of Post, Rule for Finding. . . 90
Area of Round Post, Rule 90
Area of Surface, Sliding Rupture,
Rule 88
Arithmetical Progression (Alge-
bra) 432
Astragal, or Bead, Classic Mould-
ing 323
Athens, Parthenon, Columns of. . 48
Attic, a Small Order, Top of
Building 15
INDEX.
659
Attic Story, Upper Story 15
Axes of Ellipse (Geometry) . . 585
Axiom Defined (Geometry) 348
Axis Defined 548
Balusters, Handrailing, Winding
Stairs , 310
Baluster, Platform Stairs, Position
of 250
Baluster in Round Rail, Winding
Stairs 313
Base, Shaft, and Capital Defined . 14
Bathing, Necessary Arrangements
for 45
Baths of Diocletian, Splendor of.. 27
Bead, or Astragal, Classic Mould-
ings 323
Beams, Bearings of, Rules for
Pressure 75
Beams, Breaking Weight on 74
Beams, Framed, Rules for Thick-
ness 130
Beams, Framed, Position of Mor-
tise 236
Beams, Headers Defined 130
Beams, Horizontal Thrust, Rule.s
for 72
Beams, Inclined, Effect of Weight
on 72
Beams, Load on, Effect of 74
Beams, Splicing 235
Beams, Tail, Defined 130
Beams, Trimmers or Carriage, De-
fined 130
Beams, Weight on, Proportion of. 130
Beams, White Pine, Table of
Weights 177
Beams, Wooden, Use of Limited. 154
Bearings for Girders 141
Binomials, Multiplication of (Al-
gebra) 409
Binomials, Square of (Algebra). . . 429
Binomials, Squaring (Algebra)... 410
Bisect an Angle (Geometry) 554
Bis'ect a Line (Geometry) 549
Blocking out Rail, Winding Stairs 301
Blondel's Method, Rise and Tread
in Stairs 242
Bottom Rail for Doors, Rule for
Width 316
Bow, Mr., On Economics and
Construction 166
Bowstring Girder, Cast -Iron,
Should not be Used 163
Brace, Length of, To Find (Geom-
etry) 579
Braces, Rafters, etc., To Find
Length 580
Braces in Roof, Rule for, Same as
Rafter 208
Breaking Weight Defined 84
Brick or Stone Buildings 37
Brick Walls, Modern 49
Bridge Abutments, Strength of.. . 227
Bridge Arches, Formation of 51
Bridge Arch-stones, Joints of .... 233
Bridges, Construction of Various. 223
Bridge, London, Age of Piles
under 229
Bridge Piers, Construction and
Sizes 228*
Bridge, Rib-built 224
Bridge, Rib, Construction of. . . . 225
Bridge, Rib, Framed, Construction
and Distance 226
Bridge, Rib, Radials of 226
Bridge, Rib, Table of Least Rise
in 224
Bridge, Rib, Rule for Area of 225
Bridge, Rib, Rule for Depth of. . . 226
Bridge, Roadway, Width of 227
Bridge, Stone, Arch Construction 230
Bridge, Stone, Arch-stones, Table
of Pressures on 230
Bridge, Stone, Arch, Centres for,
Bad Construction 229
Bridge, Arch, Spring of 247
Bridge, Stone, Strength of Truss-
ing 232
Bridge, Weight, Greatest on! 225
Bridge, without Tie-Beam 224
Bridging, Cross-, Additional
Strength by 137
Bridging, Cross-, Defined 137
Bridging, Cross-, Resistance by
Adjoining Beams 139
66o
INDEX.
Building, Antiquity of 5
Building, Elementary Parts of a.. 46
Building, Expression in a 35
Building by the Greeks 35
Building, Modes of, Defined 9
Building by the Romans 26
Building, Style of, Selected to Suit
Destination 35
Butt-joint on Handrail to Stairs. . 303
Butt-joint, Handrail, Stairs, Posi-
tion of 307
Byzantine Style, Lombard 10
Campanile, or Leaning Tower,
Twelfth Century 12
Capital, Uppermost Part of a
Column 15
Carriage Beam, Well-Hole in Mid-
dle, Find Breadth 136
Carriage Beam, One Header, Rule
for Breadth 133
Carriage Beam or Trimmer De-
fined rso
Carriage Beam, Rule for Breadth. 132
Carriage Beam, Two Sets of Tail
Beams, Rule for Breadth 134
Caryatides, Description and Ori-
. gin of 26
Cast-Iron Bowstring Girder,
Should not be Used 163
Cast-Iron Girder, Load at Middle,
Size of Flanges. 162
Cast-Iron Girder, Load Uniform,
Size of Flanges 163
Cast-iron Girder, Manner of Mak-
ing a 161
Cast-Iron Girder, Proper Form... 161
Cast-iron, Tensile Strength of. ... 161
Cast-Iron Untrustworthy 161
Catenary Arch, Hooke's Theory of. 51
Cathedral of Cologne n
Cathedrals, Domes of 53
Cathedrals of Pisa, Erection in
1016 12
Cavern, The Original Place of
Shelter 13
Cavetto or Cove, Classic Moulding 323
Cavetto, Grecian Moulding 327
Cavetto, Roman Moulding 329
Ceiling, Cracking, How to Pre-
vent 125
Centre of Circles, To Find (Ge-
ometry).... 556
Centre of Gravity, Position of 71
Centre of Gravity, Rule for Find-
ing, Examp^es 71
Chimneys, How Arranged, 42
Chinese Structure, The Tent the
Model of . . *. 14
Chord of Circle Defined 547
Chords Giving Equal Rectangles. 363
Circle, Arc, Rule for Length of. . . 475
Circle, Area, Circumference, etc.,
Examples 652
Circle, Area, Rule for, Length of
Arc Given 478
Circle, Area, To Find 475
Circle> Circumference, To Find . . 473
Circle Defined 546
Circle, Describe within Triangle.. 566
Circle, Diameter and Circumfer-
ence 472
Circle, Diameter and Perpendicu-
lar 468
Circle Equal Given Circles, To
Make 580
Circle, Ordinates, Rule for 471
Circle, Radius from Chord and
Versed Sine 469
Circle, Sector, Area of 476
Circle, Segment, Area of 477
Circle, Segment from Ordinates. . 470
Circle, Segment, Rule for Area of. 479
Circles, Table of 649-652
Circle through Given Points 559
Circular Headed Doors 320
Circular Headed Doors, To Form
Soffit 321
Circular Headed Windows 320
Circular Headed Windows, To
Form Soffit 321
Circular Stairs, Face Mould for (i). 282
Circular Stairs, Face Mould for (2). 285
Circular Stairs, Face Mould for (3). -287
Circular Stairs, Face Mould, First
Section 283
INDEX.
66l
Circular Stairs, Falling Mould for
Rail 281
Circular Stairs, Handrailing for . . 278
Circular Stairs, Plan of 279
Circular Stairs, Plumb Bevel De-
fined 282
Circular Stairs, Timbers Put in
after Erection *. . . 253
Cisterns, Wells, etc., Table of
Capacity of 653
City Houses, General Idea of.. ... 42
City Houses, Arrangements for.. . 37
Civil Architecture Defined 5
Classic Architecture, Mouldings
in 323
Classic Moulding, Annulet or Fil-
let 323
Classic Moulding, Astragal or
Bead 323
Classic Moulding, Cavetto or
Cove 323
Classic Moulding, Cyma-Recta. . . 324
Classic Moulding, Cyma-Reversa. 324
Classic Moulding, Ogee 324
Classic Moulding, Ovolo 323
Classic Moulding, Scotia 323
Classic Moulding, Torus 323
Coffer Walls 49
Cohesive Strength of Materials. . . 76
Collar Beam in Truss 238
Cologne, Cathedral of u
Columns, Antique, Form of 48
Column, Base, Shaft and Capital. 14
Columns, Egyptian, Dimensions,
etc 33
Column, Gothic Pillar, Form of.. 48
Column, Outline of. 47
Columns, Parthenon at Athens,
Forms of 48
Column or Pillar 47
Column, Resistance of 47
Column, Shaft, Form of. 47
Column, Shaft, Swell of, Called
Entasis 48
Complex, or Ground Vault 52
Composite Arch of Titus 28
Composite, Corinthian or Roman
Order.. 28
Compression, Resistance to 77
Compression, Resistance to Crush-
ing and Bending.*. 85
Compression, Resistance to, Pres-
sures Classified 83
Compression, Resistance to, Table
of 79
Compression, Resistance to, in
Proportion to Depth 101
Compression at Right Angles and
Parallel to Length 206
Compression of Stout Posts 89
Compression and Tension,
Framed Girders 174
Compression Transversely to Fi-
bres 86
Cone Defined 548
Conic Sections 584
Conjugate Axis Defined 548
Conjugate Diameters to Axes of
Ellipse 487
Construction Essential 56
Construction of Floors, 'Roof,
etc., Economy Important 123
Construction, Framing, Heavy
Weight 56
Construction, Joints, Effect of
Many 123
Construction, Object of Defined. . 123
Construction, Simplest Form Best. 123
Construction, Superfluous Mate-
rial 56
ontents, Table of, General. . .613-624
orinthian Capital, Fanciful Ori-
gin of. 24
orinthian Order Appropriate in
Buildings.. . . . 24
orinthian Order, Character of.... 16
orinthian Order, Description of. 23
orinthian Order, Elegance of. ... 23
orinthian Order, The Favorite
at Rome 27
orinthian Order, Grecian Origin
of 16
orinthian Order, Modification of. 27
Cornice, Angle Bracket, To Ob-
tain the 343
Cornice, Eaves, To Find Depth of. 335
662
INDEX.
Cornice, Mouldings, Depth of. . . . 342
Cornice, Projection, To Find .... 342
Cornice, Projecting Part of En-
tablature 15
Cornice, Rake and Level Mould-
ings, To Match 344
Cornice, Shading, Rule for . 6n
Cornice, Stucco, for Interior, De-
signs 340
Corollary Defined (Geometry) 348
Corollary of Triangle and Right
Angle 355
Cottage Style, English 35
• Country-Seat, Style of a 37
Cross-Bridging, Additional
Strength by 137
Cross-Bridging, Furring Impor-
tant. .;..• 137
Cross-Bridging, Resistance of Ad-
joining Beams ... 139
Cross- or Herring-Bone Bridging
Defined 137
Cross-Furring Denned 125
Cross-Strains, Resistance to. . . .77, 99
Crushing and Bending Pressure.. 85
Crushing, Liability of Rafter to. . . 205
Crushing Strength of Stout Posts. 89
Cube Root, Examples in 645
Cubes, Squares and Roots, Table
of 638-645
Cubic Feet to Gallons, To Reduce. 653
Cupola or Dome 53
Curb or Mansard Roof 54
Curve Ellipse, Equations to 482
Curve Equilibrium of Dome 218
Cylinder, Defined 549
Cylinder, Platform Stairs 248
Cylinder, Platform Stairs, Lower
Edge of 249
Cylinders and Prisms, Stair-Build-
ing 257
Cyma-Recta, Classic Moulding... 324
Cyma-Reversa, Classic Moulding. 324
Cyma-Recta, Grecian Moulding . . 327
Cyma-Reversa, Grecian Moulding. 328
Deafening, Weight per Foot 177
Decagon, Defined 546
Decimals, Reduction of, Examples 647
Decorated Style, i4th Century. ... u
Decoration, Attention to be given
to.. .: 46
Decoration, Roman 27
Deflection, Defined 112
Deflection, Differs in Different Ma-
terials 113
Deflection, Elasticity not Dimin-
ished by 112
Deflection, Floor-beams, Dwell-
ings, Dimensions 127
Deflection, Floor-beams, First-
class Stores, Dimensions 128
Deflection, Floor-beams, Ordinary
Stores, Dimensions 127
Deflection, Lever, Principle of. . . 119
Deflection, Lever and Beam, Rela-
tion Between 119
Deflection, Lever, To Find, Load
at End 120
Deflection, Lever, Breadth or
Depth, Load at End 121
Deflection, Lever, Load Uniform. 121
Deflection, Lever, Breadth or
Depth, Load Uniform 122
Deflection, Lever, for Certain,
Load Uniform 122
Deflection, Load Uniform or at
Middle, Proportion of 116
Deflection, Load Uniform, Breadth
and Depth 117
Deflection, Load Uniform or at
Middle, Proportion of 119
Deflection, in Proportion to
Weight '. . 112
Deflection, Resistance to, Rule
for 113
Deflection, Safe Weight for Pre-
vention.. no
Deflection, Weight at Middle,
Breadth and Depth 114
Deflection, Weight at Middle, for
Certain 114
Deflection, Weight at Middle, Cer-
tain, for 116
Deflection, Weight Uniform, for
Certain , 117
INDEX.
Deflection, Weight Uniform, Cer-
tain, for
Denominator, Least Common (Al-
gebra) '.
Denominator, Least Common
(Fractions)
Dentils, Teeth-like Mouldings in
Cornice
Diagonal Crossing Parallelogram
(Geometry)
Diagonal of Square Forming Oc-
tagon
Diagram of Forces, Example....
Diameter, Circle, Denned
Diameter, Ellipse, Denned
Diastyle, Explanation of the Word
Diastyle, Intercolumniation
Diocletian, Baths, Splendor of. ...
Division, Fractions. Rule for
Division, by Factors (Fractions)..
Division, Quotient (Algebra)
Division, Reduction (Algebra)...
Dodecagon, Denned
Dodecagon, To Inscribe
Dodecagon, Radius of Circles
(Polygons)
Dodecagon, Side and Area (Poly-
gons)
Dome, Abutments, Strength of. . .
Domes of Cathedrals
Dome, Character of '.
Dome, Construction and Form
118
404
384
357
1 66
547
549
19
20
27
389
381
419
Dome, Small, over Stairways, Form
of 220
Dome, Spherical, To Form 221
Dome, St. Paul's, London 54
Dome, Strains on, Tendencies of. 219
Domes, Wooden 54
Doors, Circular Head 320
Doors, Circular Head, to Form
Soffit 321
Doors, Construction of 317
Doors, Folding and Sliding, Pro-
portions 316
Doors, Front, Location of 320
Doors, Height, Rule for, Width
Given 315
Door Hanging, Manner of 317
Doors, Panel, Bottom and Lock
Rail, Width 316
Doors, Panel, Four Necessary. . . 317
Doors, Panel, Mouldings, Width. 317
Doors, Panel, Styles and Muntins,
419 Width 316
546 Doors, Panel, Top Rail, Width. .. 317
569 Doors, Stop for. How to Form... 317
Doors, Single and Double, height
452 of 316
Doors, Trimmings Explained. ... 317
453 I Doors, Uses and Requirements of 315
Doors, Width of 315
of.
216
Dome, Construction and Strength
of
Dome, Cubic Parabola computed
Dome or Cupola, the
Dome, Curve of Equilibrium, rule
for
Dome, Halle du Bled, Paris
Dome, Pantheon at Rome
Dome, Pendentives of
Dome, Polygonal, Shape of An-
gle Rib
Dome, Ribbed, Form and Con-
struction
Dome, Scantling for, Table of
Thickness
53
219
53
218
54
53
53
223
217
218
Doors, Should not be Winding. .. 317
Doors, Width and Height, Propor-
tion of 315
Doors, Width, Rule for, Height
Given 316
Doric Order, Character of 16
Doric Order, Grecian Origin of. . 16
Doric Order, Modified by the Ro-
mans 27
Doric Order, Used by Greeks only
at First 19
Doric Order. Peculiarities of 17
Doric Order, Rudeness of 30
Doric Order, Specimen Buildings
in 19
Doric Temples, Fanciful Origin
of 17
Doric Temples 19
Drawing, Articles Required 536
664
INDEX.
Drawing-board Better without
Clamps 537
Drawing-board Liable to Warp,
How Remedied 537
Drawing-board, Difficulty in
Stretching Paper 539
Drawing-board, Ordinary Size 536
Drawing, Diagrams aid Under-
standing 536
Drawing, Inking in 542
Drawing, Laying Out the 541
Drawing, the Paper 537
Drawing in Pencil, To Make Lines 542
Drawing, Secure Paper to Board. 537
Drawing, Shade Lining 543
Drawing, Stretching Paper 537
Durability in a Building 37
Dwelling, Arrangement of Rooms 38
Dwellings, Floor-beams, To Find
Dimensions 127
Dwellings, Floor-beams, Safe
Weight for 126
Dwelling-houses, Dimensions and
Style •. 37
Eaves Cornice, Designs for 335
Eaves Cornice, Rule for Depth... 335
Ecclesiastical Architecture, Point-
ed Style ii
Ecclesiastical Style, Origin of. ... 14
Echinus, Grecian Moulding 327
Economy, Construction Floors,
Roofs, Bridges 123
Eddystone and Bell Rock Light
House 48
Egyptian Architecture 30
Egyptian Architecture, Appropri-
ate Buildings for 33
Egyptian Architecture, Character
of.
33
Egyptian Architecture, Origin in
Caverns 14
Egyptian Architecture. Principal
Features of 30
Egyptian Columns, Dimensions
and Proportions 33
Egyptian Walls, Massiveness of. . 33
Egyptian Works of Art 30
Elasticity of Materials 84
Elasticity not Diminished by De-
flection 112
Elasticity, Result of Exceeding
Limit 120
Elevation, a Front View 37
Elevated Tie-beam Roof Truss
Objectionable 214
Ellipse, Area 488
Ellipse, Axes, Two, To Find, Di-
ameter and Conjugate Given. .. 593
Ellipse Defined 481
Ellipse, Equations to the Curve. . 482
Ellipse, Major and Minor Axes
Defined : .. 481
Ellipse, Ordinates, Length of ... 491
Ellipse, Parameter and Axis, Re-
lation of 485
Ellipse, Practical Suggestions.... 489
Ellipse, Semi-major, Axis Defined 486
Ellipse, Subtangent Defined 486
"Ellipse, Tangent to Axes, Rela-
tion of 485
Ellipse, Tangent with Foci, Rela-
tion of 487
Ellipsis, Axes of, To Find (Geom-
etry) 585
Ellipsis, Conjugate Diameters (Ge-
ometry) 593
Ellipsis Defined 548, 585
Ellipsis, Diameter Defined 549
Ellipsis, Foci, To Find 586
Ellipsis, by Intersecting Arcs. . . . 590
Ellipsis, by Intersecting Lines... 588
Ellipsis, by Ordinates 588
Ellipsis, Point of Contact with
Tangent, To Find 593
Ellipsis, Proportionate Axes, to
Describe with 594
Ellipsis, Trammel, to Find, Axes
Given 586
Elliptical Arch, Joints, Direction
of 233
English Architecture, Early n
English Cottage Style Extensive-
ly Used 35
j England and France, Fourteenth
Century 12
INDEX.
665
Entablature, above Columns and
Horizontal 14
Entasis, Swell of Shaft of Column 48
Equal Angles Defined 349
Equal Angles, Example in 350
Equal Angles, in Circle 358
Equal Angles (Geometry) 553
Equilateral Rectangle, to De-
scribe 568
Equilateral Triangle Defined (Ge
ometry) 545
Equilateral Triangle, to Construct
(Geometry) 568
Equilateral Triangle, to Describe
(Geometry) 566
Eqilateral Triangle, to Inscribe
(Geometry) 569
Equilateral Triangle (Polygons).. 445
Eustyle Defined 20
Exponents, Quantities with Nega-
tive (Algebra) 423
Extrados of an Arch 52
Face Mould, Accuracy of, Wind-
ing Stairs 295
Face Mould, Curves Elliptical,
Winding Stairs 301
Face Mould, Drawing of, Winding
Stairs 296
Face Mould, Sliding of, Winding
Stairs 299
Face Mould, Application of, Plat-
form Stairs 275
Face Mould, a Simple, Kell's
Method for 268
Factors, Multiplication (Algebra) 409
Factors, Two, Squaring Difference
of (Algebra) 412
Fibrous Structure of Materials.. . 76
Figure Equal, Given Figure (Ge-
ometry) 575
Figure, Nearly Elliptical, To Make
(Geometry) 591
Fillet or Amulet, Classic Mould-
ing 323
Fire-proof Floors, Action of Fire
on JJ3
Flanges, Cast-iron Girder 163
Flanges, Area of, Tubular Iron
Girder 155
Flanges, Area of Bottom, Tubular
Iron Girder 159
Flanges, Load at Middle of Cast-
iron Girder, Sizes 162
Flanges, Load Uniform on Tubu-
lar Iron Girder, Sizes 156
Flanges, Proportion of, Tubular
Iron Girder 157
Flexure, Compared with Rup-
ture 84
Flexureof Rafter 205
Flexure, Resistance to, Defined. . 145
Floor-arches, How Constructed. . 153
Floor-arches, Tie-rods, Dwellings,
Sizes 153
Floor-arches, Tie-rods, First-class
Stores, Sizes 153
Floor-beams, Distance from Cen-
tres, Sizes Fixed 129
Floor-beams, Dwellings, Safe
Weight for 126
Floor-beams, Dwellings, Deflec-
tion Given, Sizes 127
Floor-beams, First-class Stores,
Deflection Given, Sizes 128
Floor-beams, Ordinary Stores, De-
flection Given, Sizes 127
Floor-beams, Stores, Safe Weight
for 126
Floor-beams, Reference to Rules
for Sizes 125
Floor-beams, Reference to Trans-
verse Strains 126
Floor-beams, Proportion of
Weight on All 130
Floors Constructed, Single or
Double 124.
Floors, Fire-proof Iron, Action of
Fire on 143
Floors, Framed, Seldom Used. .. 124
Floors, Framed, Openings in 130
Floors, Headers, Defined 130
Floofs, Ordinary, Effect of Fire
on 143
Floors, Solid Timber, Dwellings
and Assembly, Depth 143
666
INDEX.
Floors, Solid Timber, First-class
Stores, Depth 144
Floors, Solid Timber, to Make
Fire-proof 143
Floors, Tail-beams Defined 130
Floors, Trimmers or Carriage-
beams Defined 130
Floors, Wooden, More Fire-proof
than Iron, Some* Cases 143
Flyers and Winders, Winding
Stairs 251
Foci Defined 548
Foci of Ellipsis To Find 586
Foci of Ellipse, Tangent 487
Force Diagram, Load on Each
Support 179
Force Diagram, Truss, Figs. 59,
68 and 69 179
Force Diagram, Truss, Figs. 60, 70
and 71 180
Force Diagram, Truss, Figs. 61,
72 and 73 181
Force Diagram Truss, Figs. 63
74 and 75 183
Force Diagram, Truss, Figs. 64,
77 and 78 184
Force Diagram, Truss, Figs. 65,
78 and 79 185
Force Diagram, Truss, Figs. 66,
8oand8i 186
Forces, Parallelogram of 59
Forces, Composition of 66
Forces, Composition, Reverse of
Resolution 67
Forces, Resolution of 59
Forces, Resolution of, Oblique
Pressure 59
Foundations, Description of 47
. Foundations in Marshes, Timbers
Used 47
Fractions, Addition, Like Denom-
inators 382
Fractions Added and Subtracted
(Algebra) 403
Fractions Changed by Division!. 380
Fractions Defined 378
Fractions, Division, Rule for 389
Fractions, Division by Factors... 381
Fractions Divided Graphically. . . 388
Fractions Graphically Expressed. 378
Fractions, Improper, Defined.... 380
Fractions, Least Common Denom-
inator , 384
Fractions, Multiplication, Rule. . . 387
Fractions Multiplied Graphically. 386
Fractions, Numeratorand Denom-
inator 378
Fractions, Reduce Mixed Num-
bers 381
Fractions, Reduction to Lowest
Terms 384
Fractions Subtracted (Algebra). . . 405
Fractions, Subtraction Like De-
nominators 383
Fractions, Unlike Denominators
Equalized 383
Framed Beams, Thickness of,
Rules 130
Framed Girder, Bays Defined 167
Framed Girders, Compression and
Tension, Dimensions 174
Framed Girders, Construction
and Uses 166
Framed Girders, Height and
Depth 167
Framed Girders, Kinds of Pres-
sure 173
Framed Girders, Long, Construc-
tion of '. 174
Framed Girders, Panels on Under
Chord, Table of 167
Framed Girders, Ties and Struts,
Effect of 174
Framed Girders, Triangular Pres-
sure, Upper Chord 168
Framed Girders, Triangular Pres-
sure, Both Chords 171
Framed Openings in Floors 130
Framing Beams, Effect of Splic-
ing 235
Framing Roof Truss 237
Framing Roof Truss, Iron Straps,
Size of 239
France and England, Fourteenth
Century 12
Friction, Effect of 82
INDEX.
667
Frieze between Architrave and
Cornice 15
Furring Denned 125
Gable, a Pediment in Gothic Ar-
chitecture 15
Gaining a Beam Denned 100
General Contents, Table of 613-624
Geometrical Progression (Alge-
bra)- 435
Geometry, Angles of Triangle,
Three, Equal Right Angle 354
Geometry Chords Giving Equal
Rectangles 363
Geometry Denned 544
Geometry, Divide a Given Line. . 555
Geometry, Divisions in Line Pro-
portionate 583
Geometry, Elementary 347
Geometry, Equal Angles 553
Geometry, Equal Angles, Ex-
ample 350
Geometry, Figure Equal to Given
Figure, Construct 575
Geometry, Figure Nearly Ellipti-
cal by Compasses 591
Geometry, Measure an Angle. . . . 348
Geometry Necessary in Handrail-
ing, Stairs 257
Geometry, Opposite Angles Equal. 354
Geometry, Parallel Lines 555
Geometry, a Perpendicular, To
Erect 550
Geometry.Perpendicular.let Fall a. 551
Geometry, Perpendicular, Erect at
End of Line 551
Geometry, Perpendicular, Let Fall
Near End of Line 553
Geometry, Plane Denned (Stairs). 257
Geometry, Point of Contact 558
Geometry, Points, Three Given,
Find Fourth 559
Geometry, Right Line Equal Cir-
cumference 566
Geometry, Right Lines, Propor-
tion Between 584
Geometry, Right Lines, Two
Given, Find Third 582
Geometry, Square Equal Rec-
tangle, To Make 581
Geometry, Square Equal Given
Squares, To make 577
Geometry, Square Equal Triangle,
To Make .. 582
German or Romantic Style, Thir-
teenth and Fourteenth Centuries, n
Girder, Bearings, Space Allowed
for 141
Girder, Bow-String, Cast-Iron,
Should not be Used 163
Girder, Bow-String, Substitute for. 163
Girder, Construction with Long
Bearings 140
Girder, Cast-Iron, Load Uniform,
Flanges 163
Girder, Cast-Iron, Load at Middle,
Flanges 162
Girder, Cast-Iron, Proper Form
of 161
Girder Denned, Position and Use
of 140
Girder, Different Supports for. . . . 140
Girder, Dwellings, Sizes for 141
Girder, Framed, Bays Denned 167
Girders, Framed, Compression
and Tension, Dimensions 174
Girder, Framed, Construction of.. 140
Girder, Framed, Construction and
Uses 166
Girder, Framed, Construction of
Long 174
Girder, Framed, Kinds of Pres-
sure 173
Girders, Framed, Height and
Depth 167
Girders, Framed, Panels on Under
Chord, Table of. 167
Girders, Framed, Triangular Pres-
sure Upper Chord 108
Girders, Framed, Triangular Pres-
sure Both Chords 171
Girders, Framed, and Tubular
Iron 140
Girders, First-Class Stores, Sizes
for 141
Girders, Sizes, To Obtain 141
668
INDEX.
Girders, Strengthening, Manner
of. 140
Girders, Supports, Length of, Rule. 157
Girders, Tubular Iron, Construc-
tion of 154
Girders, Tubular Iron, Area of
Flange. Load at Middle 154
Girders, Tubular Iron, Area of
Flange, Load at any Point 155
Girders, Tubular Iron, Area of
Flange, Load Uniform 156
Girders, Tubular Iron, Dwellings,
Area of Bottom Flange 159
Girders, Tubular Iron, First-Class
Stores, Area of Bottom Flange.. 160
Girders, Tubular Iron, Rivets, Al-
lowance for 157
Girders, Tubular Iron, Flanges,
Proportion of 157
Girders, Tubular Iron, Shearing
Strain 15?
Girders, Tubular Iron.Web, Thick-
ness of i58
Girders, Weakening, Manner of. . 140
Girders, Wooden, Objectionable. 154
Girders, Wooden, Supporting,
Manner of 154
Glossary of Terms 627-637
Gothic Arches 51
Gothic Buildings, Roofs of 55
Gothic and Norman Roofs, Con-
struction of 178
Gothic Pillar, Form of 48
Gothic Style, Characteristics of. . . 12
Goths, Ruins Caused by 12
Granular Structure of Materials. . 76
Gravity, Centre of, Position 71
Gravity, Centre of, Examples, and
Rule for 71
Grecian Architecture, History of. 6
Grecian Art, Elegance of 27
Grecian Moulding, Cyma-Recta.. 327
Grecian Moulding, Cyma-Re-
versa. . . -. 328
Grecian Moulding, Echinus and
Cavetto 327
Grecian Moulding, Scotia 326
Grecian Moulding, Torus 326
Grecian Orders Modified by the
Romans 27
Grecian Origin of the Doric Or
der 16
Grecian Origin of Ionic Order ... 16
Grecian Style in America 13
Grecian Styles, their Different
Orders 16
Greek Architecture, Doric Order
Used 19
Greek Building 35
Greek Moulding, Form of 325
Greek, Persian, and Caryatides
Orders 24
Greek Style Originally in Wood.. 14
Greek Styles Only Known by
Them 16
Groined or Complex Vault 52
Halle du Bled, Paris, Dome of. . . 54
Halls of Justice, N. Y. C., Speci-
men of Egyptian Architecture. . 8
Handrailing, Circular Stairs 278
Handrailing, Platform Stairs. ... 269
Handrailing, Platform Stairs, Face
Mould 264
Handrailing, Platform Stairs,
Large Cylinder 271
Handrailing Stairs, Geometry
Necessary 257
Handrailing Stairs," Out of Wind"
Defined. 257
Handrailing Stairs, Tools Used. . 257
Handrailing, Winding Stairs. .256, 289
Handrailing Winding Stairs, Bal-
usters Under Scroll 310
Handrailing, Winding Stairs,
Centres in Square 308
Handrailing, Winding Stairs, Face
for Scroll 311
Handrailing, Winding Stairs, Fall-
ing Mould 310
Handrailing, Winding Stairs, Gen-
eral Considerations 258
Handrailing, Winding Stairs,
Scroll for 308
Handrailing, Winding Stairs,
Scroll at Newel 309
INDEX.
669
Handrailing, Winding Stairs,
Scroll Over Curtail Step 309
Handrailing, Winding Stairs,
Scroll for Curtail Step 310
Headers, Breadth of. 130
Headers Defined 130
Headers, Mortises, Allowance for
Weakening by 131
Headers, Stores and Dwellings,
Same for Both 132
Hecadecagon, Complete Square
(Polygons) 458
Hecadecagon, Radius of Circles
(Polygons) 455
Hecadecagon, Rules (Polygons). . 459
Hecadecagon, Side and Area
(Polygons) 457
Height and Projection, Numbers
of an Order 16
Hemlock, Weight per Foot Super-
ficial 177
Heptagon Defined 546
Herring-bone Bridging Defined... 137
Hexagon Defined 546
Hexagon, To Inscribe 569
Hexagons, Radius of Circles 447
Hexastyle, Intercolumniation. . . . 20
Hindoo Architecture, Ancient,
Character of 30
Hip-Rafter, Backing of 216
Hip-Roofs, Diagram and Expla-
nation 215
History of Architecture 44
Hogged Ridge in Roof Truss 238
Homologous Triangles (Geom-
etry) 362
Homologous Triangles (Ratio
and Proportion) 370
Hooke's Theory of an Arch 50
Hooke's Theory, Bridge Arch,
Pressure on. ... 51
Hooke's Theory, Catenary Arch. . 51
Horizontal and Inclined Roofing,
Weight 190
Horizontal Pressure on Roof, To
Remove 74
Horizontal Thrust in Beams 72 i
Horizontal Thrust, Tendency of. . 88 i
Hut, Original Habitation 13
Hydraulic Method, Testing
Woods 80
Hyperbola Defined 548, 585
Hyperbola, Height, To Find, Base
and Axis Given 585
Hyperbola by Intersecting Lines. 595
Hypothenuse, Equality of Squares
(Algebra) 416
Hypothenuse, Formula for (Trig-
onometry) 516
Hypothenuse, Side, To Find (Ge-
ometry) 579
Hypothenuse, Right Angled Tri-
angle (Geometry) 355
Hypothenuse, Triangle (Trigo-
nometry) 518
Ichnographic Projection, Ground
Plan 37
Improper Fractions Defined 380
India Ink in Drawing 540
Inertia, Moment of, Defined 145
Inking-in Drawing . . 542
Inside Shutters for Windows, Re-
quirements 319
Instruments in Drawing 540
Intercolumniation Defined 17
Intercolumniation of Orders 20
Intrados of Arch 52
Ionic Order, Character of. 16
Ionic Order, Grecian Origin of. . . 16
Ionic Order Modified by the Ro-
mans. ...i 27
Ionic Order, Origin of 20
Ionic Order, Suitable for What
Buildings 20
Ionic Volute, To Describe an. ... 20
Iron Beams, Breaking Weight at
Middle 148
Irpn Beams, Deflection, To Find,
Weight at Middle 147
Iron Beams, Deflection, To Find,
Weight Uniform 150
Iron Beams, Dimensions, To Find,
Weight any Point 149
Iron Beams, Dimensions, To Find,
Weight Uniform 149
6 ;o
INDEX.
Iron Beams, Dwellings, Distance
from Centres 151
Iron Beams, First-Class Stores,
Distance from Centres 152
Iron Beams, Rectangular Cross-
Section 145
Iron Beams, Rolled, Sizes 145
Iron Beams, Safe Weight, Load
any Point 148
Iron Beams, Safe Weight, Load
Uniform 151
Iron Beams, Table IV 146
Iron Beams, Weight at Middle,
Deflection Given 146
Iron Fire-Proof Floors, Action of
Fire On 143
Iron Straps, Framing, to Prevent
Rusting 239
Irregular Polygon, Trigon (Geom-
etry) 546
Isosceles Triangle Denned 545, 584
Italian Architecture, Thirteenth,
Fourteenth, and Fifteenth Cen-
turies 12
Italian Use of Roman Styles 13
Italy, Tuscan Order the Principal
Style 30
Jack-Rafters, Location of 212
jack-Rafters and Purlins in Roof. 211
Jack-Rafters, Weight per Superfi-
cial Foot 189
Joists and Studs Defined 174
Jupiter, Temple of, at Thebes, Ex-
tent of 33
Kell's Method, Simple Face
Mould, Stairs 268
Keystone for Arch, Position of. ... 50
King-Post, Bad Framing, Effect
of ... 237
King-Post, Location of 213
King-Post in Roof 54
Lamina in Girders Defined 174
Lancet Arch 51
Lateral Thrust in Arch 52
Laws of Pressure 57
Laws of Pressure, Inclined, Ex-
amples 57
Laws of Pressure, Vertical, Exam-
ples 57
Leaning Tower or Campanile,
Twelfth Century 12
Length, Breadth, or Thickness,
Relation to Pressure 78
Lever, Breadth or Depth, To
Find in
Lever, Deflection as Relating to
Beam 1 19
Lever, Deflection, Load at End.. 120
Lever, Deflection, Load Uniform. 121
Lever, Deflection, Breadth or
Depth, Load at End 121
Lever, Deflection, Breadth or
Depth, Load Uniform 122
Lever, Deflection, Load Required. 122
Lever Formula, Proportionals in
(Algebra) 421
Lever Load Uniformly Distrib-
uted in
Lever, Load at One End no
Lever Principle Demonstrated
(Ratio) 375
Lever, Support, Relative Strength
of One no
Light-Houses, Eddystone and Bell
Rock 48
Line Defined (Geometry) 544
Lines, Divisions in, Proportionate
(Geometry) 583
Lintel, Position of 49
Lintel, Strength of 49
Load, per foot, Horizontal 192
Load on Roof Truss, per Superfi-
cial Foot 189
Load on Tie-Beam, Ceiling, etc. . 190
Lock Rail for Doors, Width 316
Logarithms Explained (Algebra).. 425
Logarithms, Examples 426
Logarithms, Sine and Tangents
(Polygons) 464
Lombard, Byzantine Style 10
Lombard Style, Seventh Century. 10
London Bridge, Piles, Age of 229
INDEX.
67I
Materials, Cohesive Strength of. . 76
Materials, Compression, Resist-
ance to. 77
Materials, Cross-strain, Resistance
to 77
Materials, Structure of 76
Materials, Tension, Resistance to. 77
Materials Tested, General De-
scription 80
Materials, Weights, Table of. . . . 654
Major and Minor Axes of Ellipse
Denned 481
Marshes, Foundation for Timbers
in 47
Mathematics Essential 347
Maxwell, Prof. I. Clerk, Diagrams
of Forces, etc 165
Memphis, Pyramids of, Estimate
of Stone in 33
Minster, Tower of Strassburg n
Minutes, Sixty Equal Parts, to
Proportion an Order 15
Mixed Numbers in Fractions, To
Reduce 381
Modern Architecture, First Ap-
pearance of 9
Modern Tuscan, Appropriate for
Buildings 30
Moment of Inertia Defined 145
Mono-triglyph, Explanation of the
Word 19
Monuments, Ancient, Their Archi-
tects 6
Moorish and Arabian Styles, An-
tiquities of ii
Mortises, Proper Location of — . 100
Mortising, Beam, Effect on
Strength of 100
Mortising Beam at Top, Injurious
Effect of 100
Mortising Beam, Effect of 231
Mortising, Beam, Position of 236
Mortising Headers, Allowance for
Weakening 131
Moulding, Classic, Astragal or
Bead 323
Moulding, Classic, Annulet or
Fillet 323
Mouldings, Classic Architecture. 323
Moulding, Classic, Cavctto or
Cove 323
Moulding, Classic, Cyma-Recta. . 324
Moulding, Classic, Cyma-Reversa. 324
Moulding, Classic, Ogee 324
Moulding, Classic, Ovolo 323
Moulding, Classic, Scotia 323
Moulding, Classic, Torus 323
Mouldings, Common to all Or-
ders 324
Mouldings Defined 323
Mouldings, Diagrams of 330
Mouldings, Doors, Rule for Width. 317
Moulding, Grecian, Cyma-Recta. 327
Moulding, Grecian, Cyma-Rc-
versa 328
Moulding, Grecian Echinus and
Cavetto 327
Mouldings, Greek, Form of. 325
Mouldings, Grecian Torus and
Scotia 326
Mouldings, Modern 331
Moulding, Modern, Antae Cap... 334
Mouldings, Mbdern Interior, Dia-
grams 332
Mouldings, Modern, Plain 333
Mouldings,Names, Derivations of. 324
Mouldings, Profile Defined 326
Mouldings, Roman, Forms of. .. . 325
Mouldings, Roman, Comments on. 329
Mouldings, Roman, Ovolo and
Cavetto 329
Mouldings,Uses and Positions of. 324
Multiplication (Algebra) 408
Multiplication, Plus and Minus
(Algebra) 415
Multiplication, Three Factors (Al-
gebra) 408
Multiplication, Fractions 387
Newel Cap, Form of, Winding
Stairs 312
Nicholson's Method, Plane
Through Cylinder (Stairs) 259
Nicholson's Method, Twists in
Stairs 259
Nonagon Defined 546
672
INDEX".
Normal and Subnormal in Para-
bola 496
Norman and Gothic Construction
of Roofs 178
Norman Style, Peculiarities of. . . n
Nosing and Tread, Position in
Stairs 241
Oblique Angle Defined 544
Oblique Pressure, Resolution of
Forces . . . . 59
Oblique Triangle, Difference Two
Angles (Trigonometry) 523
Oblique Triangle, First Class
(Trigonometry) , 520
Oblique Triangles, First Class,
Formulae (Trigonometry) 531
Oblique Triangles, Second Class
(Trigonometry) 522
Oblique Triangles, Second Class,
Formulae (Trigonometry) 532
Oblique Triangles, Third Class
(Trigonometry) 526
Oblique Triangles, Third Class,
Formula; (Trigonometry) 534
Oblique Triangles, Fourth Class
(Trigonometry) 528
Oblique Triangles, Fourth Class,
Formulae (Trigonometry) 534
Oblique Triangles, Two Sides
(Trigonometry) 521
Oblique Triangles, Sines and
Sides (Trigonometry) 519
Obtuse Angle Denned 349, 544
Obtuse Angled Triangle Denned. 545
Octagon, Buttressed, Find Side
(Geometry) 571
Octagon Defined 546
Octagon, Diagonal of Square
Forming 357
Octagon, Inscribe a (Geometry). . 570
Octagon, Rules (Polygons) 451
Octagon, Radius of Circles (Poly-
gons) 449
Octastyle, Intercolumniation 20
Ogee Mouldings, Classic 324
Opposite Angles Equal (Geome-
try) 354
Order of Architecture, Three
Principal Parts 14
Orders of Architecture, Persians
and Caryatides 24
Ordinates to an Arc (Geometry). . 563
Ordinates, Circle, Rule for 471
Ordinates of Ellipse 491
Ostrogoths, Style of the 9
Oval, To Describe a (Geometry). . 591
Ovolo, Classic Moulding 323
Ovolo, Roman Moulding 329
Paper, The, in Drawing, Secure to
Board , . . 537
Pantheon at Rome, Dome of, and
Walls 53
Pantheon and Roman Buildings,
Walls of 49
Parabola, Arcs Described from... 503
Parabola, Area, Rule for 509
Parabola, Axis and Base, to find
(Geometry) 585
Parabola, Curve, Equations to... . 493
Parabola Defined 492
Parabola Defined (Geometry).. 548, 585
Parabola, Diameters 497
Parabola Described from Ordi-
nates 504
Parabola Described from Diame-
ters 507
Parabola Described from Points.. 502
Parabola of Dome Computed 219
Parabola, General Rules 499
Parabola by Intersecting Lines. . . 594
Parabola Mechanically Described. 500
Parabola, Normal and Subnor-
mal 496
Parabola, Ordinate Defined 496
Parabola, Subtangent 496
Parabola, Tangent 493
Parabola, Vertical Tangent De-
fined 495
Parabolic Arch, Direction of
Joints 234
Parallel Lines Defined 544
Parallel Lines (Geometry) 555
Parallelogram, Construct a 576
Parallelogram Defined 545
INDEX.
673
Parallelogram Equal to Triangles,
To Make.. 576
Parallelogram of Forces, Strains
by 165
Parallelograms Proportioned to
Bases (Geometry) 360
Parallelogram in Quadrangle
(Geometry) 364
Parallelogram, Same Base (Geom-
etry) 352
Parameter Defined 548
Parameter, Axes (Ellipse) 485
Parthenon at Athens, Columns of. 48
Partitions, Bracing and Trussing. 176
Partitions, How Constructed 174
Partition, Door in Middle, Con-
struction 175
Partition, Doors at End, Construc-
tion of 176
Partition, Great Strength, Con-
struction 176
Partitions, Location and Connec-
tion •. 175
Partitions, Materials, Quality of. . 175
Partitions, Plastered, Proper Sup-
ports for. 175
Partitions, Pressure on. Rules 177
Partitions, Principal, of what Com-
posed 175
Partitions, Trussing in, Effects of. 175
Pedestal, a Separate Substruc-
ture 14
Pediment, Triangular End of
Building 15
Pencil and Rulers, Drawing 540
Pentagon Defined 546
Pentagon, Circumscribed Circles
(Polygons) 463
Perpendicular Height of Roof, To
find 579
Perpendicular, Erect a 550
Perpendicular, Erect a, at End of
Line 551
Perpendicular, Let Fall a 551
Perpendicular, Let Fall a, at End
of Line 553
Perpendicular Style, Fifteenth
Century 12
Perpendicular in Triangle (Poly-
gons) 440
Persians, Origin and Description
of 24
Persians and Caryatides, Orders
Used by Greeks 24
Piers, Arrangement, in City Front
of House 44
Piers, Bridges, Construction and
Sizes 228
Piles, London Bridge, Age of. ... 229
Pine, White, Beams, Table of
Weights for. 177
Pisa, Cathedral of, Eleventh Cen-
tury 12
Pisa, Cathedral of, Erection in
1016 12
Pise Wall of France 49
Pitch Board, To Make, for Stairs. 247
Pitch Board, Winding Stairs 252
Plane Defined 257
Plane Defined (Geometry) 544
Plank.Weightof.on Roof, per foot. 189
Plastering, Defective, To what
Due 174
Plastering, Strength of 174
Plastering, Weight per foot 177
Platform Stairs, Baluster, Posi-
tion of 250
Platform Stairs Beneficial 240
Platform Stairs, Cylinder of. 248
Platform Stairs, Cylinder, Lower
Edge 249
Platform Stairs, Face Mould, Ap-
plication of Plank 273
Platform Stairs, Face Mould,
Handrailing in 264
Platform Stairs, Face Mould, Sim-
ple Method 267
Platform Stairs, Face Mould,
Moulded Rails 274
Platform Stairs, Face Mould, Ap-
plication of 275
Platform Stairs, Face Mould With-
out Canting Plank 272
Platform Stairs, Handrail to 269
Platform Stairs, Handrailing Large
Cylinder 271
674
INDEX.
PAGE I
Platform Stairs, Railing Where
Rake Meets Level 272
Platform Stairs, Twist-Rail, Cut-
ting of 277
Platform Stairs, Wreath of Round
Rail 267
Point of Contact (Geometry) 558
Point Denned (Geometry) 544
Pointed Style, Ecclesiastical Arch-
itecture ii
Polygons, Angles of 462
Polygons, Circumscribed and In-
scribed Circles, Radius of 460
Polygons Defined (Geometry). . . . 546
Polygons, Equilateral Triangle. .. 445
Polygons, General Rules 461
Polygons, Irregular, Trigon (Ge-
ometry) 546
Polygons, Perpendicular in Tri-
angle 440
Polygon, Regular, Defined (Geom-
etry) 546
Polygons, Regular, To Describe
(Geometry) 573
Polygons, Regular, To Inscribe in
Circle (Geometry) 572
Polygons, Sum and Difference,
Two Lines 439
Polygons, Table Explained 466
Polygons, Table of Multipliers. .. 465
Polygons, Triangle, Altitude of.. 442
Polygonal Dome, Shape of Angle-
Rib 223
Posts, Area, To Find 86
Posts, Diameter, To Find 92
Posts, Rectangular, Safe Weight.. 92
Posts, Rectangular.To Find Thick-
ness 94
Posts, Rectangular, Breadth Less
than Thickness 96
Posts, Rectangular, To Find
Breadth 95
Posts, To Find Side 93
Posts, Slender, Safe Weight for. . 91
Posts, Stout. Crushing Strength of. 89
Pressures Classified 85
Pressure, Oblique, Resolution of
Forces 59
Pressure, Triangular, Framed
Girders 171
Pressure, Upper Chord, Triangu-
lar Girder 168
Prisms Cut by Oblique Plane 259
Prisms and Cylinders, Stair-Build-
ing 257
Prisms Defined (Stairs) 257, 259
Prism, Top, Form of, in Perspec-
tive 259
Profile of Mouldings Defined. . . . 326
Progression, Arithmetical (Alge-
bra) 432
Progression.Geometrical (Algebra) 435
Projection and Height, Members
of Orders of Architecture 16
Protractor, Useful in Drawing. .. 541
Purlins and Jack-Rafters in Roof. 211
Purlins, Location of 212
Pyramids of Memphis, Amount of
Stone in 33
Pycnostyle, Explanation of 20
Quadrangle Defined 545
Quadrangle Equal Triangle 353
Quadrant Defined 547
Quantities, Addition and Sub-
traction (Algebra) 424
Quantities, Division of (Algebra). 424
Quantities, Multiplication of (Al-
gebra) 424
Queen-Post, Location of 213
Queen-Post in Roof 54
Radials of Rib in Bridge 226
Radials of Rib for Wedges 226
Radicals, Extraction of (Algebra). 425
Radius of Arc, To Find 561
Radius of Circle Defined 547
Rafters, Braces, etc., Length, To
Find 580
Rafters, Least Thrust, Rule for. . . 62
Rafters, Length of, To Find 578
Rafters, Liability to Crush Other
Materials 205
Rafters, Liability to Being Crushed 205
Rafters, Liability to Flexure 205
Rafters, Minimum Thrust of 62
INDEX.
6/5
PAGE
Rafters in Roof, Effect of Weight
on 179
Rafters in Roof, Strains Subjected
to , 205
Rafters and Tie-Beams, Safe
Weight 87
Rafters, Uses in Roof 54
Rake in Cornice Matched with
Level Mouldings 344
Railing, Platform Stairs Rake
Meets Level 272
Ratio or Proportion, Equals Mul-
tiplied 367
Ratio or Proportion, Equality of
Products 370
Ratio or Proportion, Equality of
Ratios 367
Ratio or Proportion Equation,
Form of 367
Ratio or Proportion, Examples.. . 366
Ratio or Proportion, Four Propor-
tionals, to Find 377
Ratio or Proportion, Homologous
Triangles 370
Ratio or Proportion, Lever Prin-
ciple in 372
Ratio or Proportion, Lever Prin-
ciple Demonstrated 375
Ratio or Proportion, Multiply an
Equation 368
Ratio or Proportion, Multiply and
Divide One Number 368
Ratio or Proportion, Rule of
Three 366
Ratio or Proportion, Steelyard as
Example in 371
Ratio or Proportion, Terms of
Quantities 367
Ratio or Proportion, Transfer a
Factor 369
Rectangle Defined 545
Rectangle, Equilateral, To De-
scribe 5°8
Rectangular Cross-Section, Iron
Beams 145
Reduction Cubic Feet to Gallons,
Rule 653
Reduction Decimals, Examples.. 647
Reflected Light, Opposite of
Shade 611
Regular Polygon in Circle, To In-
scribe (Geometry) 572
Regular Polygon Defined (Geom-
etry) 546
Regular Polygons, To Describe
(Geometry) 573
Resistance, Capability of 86
Resistance to Compression, Ap-
plication of Pressure 85
Resistance to Compression,
Crushing and Bending 85
Resistance to Compression, Mate-
rials 77
Resistance to Compression, Pres-
sure Classified 85
Resistance to Compression in
Proportion to Depth 101
Resistance to Compression, Stout
Posts, Rule 89
Resistance to Compression, Table
of Woods 79
Resistance to Cross-Strains 77
Resistance to Cross-Strains De-
fined 99
Resistance to Deflection, Rule... . 113
Resistance Depending on Com-
pactness and Cohesion 78
Resistance Depending on Loca-
tion, Soil, etc 79
Resistance to Flexure Defined. . . 145
Resistance Inversely in Propor-
tion to Length 102
Resistance to Oblique Force 206
Resistance, Power of, Hew Ob-
tained 78
Resistance, Proportion to Area. . . 86
Resistance, Strains, To What Due. 78
Resistance to Tension Greatest
in Direction of Length 81
Resistance to Tension, Proportion
in Materials 81
Resistance to Tension, Table of
Materials 82
Resistance to Tension, Materials. 77
Resistance to Tension, Results
from Transverse Strains . . 82
6;6
INDEX.
PAGE
Resistance to Transverse Strains, Roman Architecture, Ruins of. . . n
Table of. 83 Roman Architecture, Excess of
Resistance to Transverse Strains, I Emichment 46
Description of Table 84 Roman Building 26
Resistance Variable in One Ma- Roman Composite and Corinthian
terial.... 79 Orders.. . 28
Reticulated Walls . < 49 j Roman Decoration 27
Rhomboid Defined 546 Roman Empire, Overthrow of. ... 13
Romans, Ionic Order Modified by 27
Roman Moulding, Cavetto 329
Roman Mouldings, Comments on. 329
Roman Moulding, Ovolo 329
Rhombus Defined ' . . . . 545
Ribbed Bridge, Area of Rule 225
Ribbed Bridge, Built 224
Ribbed Bridge, Least Rise, Table
of 224
Right Angle Defined 348, 544
Right Angle in Semicircle (Ge-
ometry) 355
Right Angle, To Trisect a 554
Right Angled Triangle Defined . . 545
Right Angled Triangle, Squares
on (Algebra) 417
Right Angled Triangles (Trigo-
nometry) 510
Right Angled Triangles, Formula
for (Trigonometry) 530
Right Lines (Geometry) 584
Right Line Equal Circumference. 566
Right Lines, Mean Proportionals
Between 584
Right Lines, Two Given, Find
Third 582
Right Lines, Three Given, Find
Fourth 583
Right or Straight Line Defined. . . 544
Right Prism Defined (Stairs) 257
Risers, Number of, Rule to Ob-
tain (Stairs) 246
Rise and Tread (Stairs) 241
Rise and Tread, Connection of
(Stairs) 248
Rise and Tread, Blondel's Method
of Finding (Stairs) 242
Rise and Tread, Table of, for
Shops and Dwellings (Stairs). . . 245
Rise and Tread, To Obtain (Wind-
ing Stairs) 251
Rolled Iron Beams, Extensive Use
Roman Mouldings, Forms of.. ... 325
49
26
13
of.
161
Roman Architecture Defined. .... 7
Roman Pantheon, etc., Walls of. .
Roman Styles of Architecture. . . .
Roman Styles Spread by the Ital-
ians
Romantic or German Style, Thir-
teenth and Fourteenth Centu-
ries ii
Rome, Ancient Buildings of. 12
Rome and Greece, Architecture
of 8
Roof, The 54
Roofs, Ancient Norman and
Gothic, Construction 178
Roof Beams, Weight per Super-
ficial Foot 189
Roof, Brace in, Rule Same as for
Rafter 208
Roofs, Construction of. 55
Roof Covering, Mode of 188
Roof Covering, Weights, Table of. 191
Roof, Curb or Mansard 54
Roofs, Diagrams and Description
of 212
Roof, Gothic Buildings 55
Roofs, Gothic and Norman Puild-
ings, Construction 178
Roofs, Hip, Diagram and Exj
nation 215
Roof, Hip 54
Roof, Horizontal Pressure, To Re-
move from 74
Roof, Jack-Rafters and Purlins.. . 211
Roof, King-Post in 54
Roof, Load per Foot Horizontal,
Rule,
IQ2
INDEX.
677
Roof, Load, Total per Foot Hori-
zontal, Rule 197
Roofs, Modern, Trussing Neces-
sary 178
Roofs, Norman and Gothic Build-
ings, 178
Roof, Pent, To Find 54
Roof, Perpendicular Height, To
Find 579
Roof Plank, Weight per Super-
ficial Foot 189
Roof, Planning a 188
Roof, Pressure on 55
Roof, Queen-Post in 54
Roof, Rafters in 54
Roof, Sagging, To Prevent. ...... 54
Roof, Slope Should Vary Accord-
ing to Climate 191
Roof Supports. Distance between. 189
Roof, Suspension Rods, Safe
Weight for 210
Roof, Tie-Beam in 54
Roof, Tie-Beam, Tensile Strain,
Rule ,. 204
Roof Timbers, Mortising 55
Roof Timbers, Scarfing of 55
Roof Timbers, Splicing of 55
Roof Timbers, Strains by Parallel-
ogram of forces 198
Roof Timbers, Strain Shown Ge-
ometrically 199, 202
Roof Truss, Arched Ceiling 214
Roof Truss, Elevated Tie-Beam
Objectionable 214
Roof Truss, Elevating Tie-Beam,
Effect of 187
Roof Truss, Force Diagram, Figs.
59, 68, and 69 179
Roof Truss, Force Diagram, Figs.
60, 70, and 7 r 180
Roof Truss, Force Diagram, Figs.
61, 72, and 73 181
Roof Truss, Force Diagram, Figs.
63,74, and 75 183
Roof Truss, Force Diagram, Figs.
64, 77, and 78 184
Roof Truss, Force Diagram, Figs.
66, 80, and 81 186
Roof Truss, Load on 189
Roof Trusses, Strains, Effect of, on
Different 179
Roof Truss, Weights, Table of, per
Superficial Foot 189
Roof Truss, Weight per Superfi-
cial Foot 190
Roof, Trussing in 54
Roof Trussing, Designs for 178
Roof Trussing, Framing for 237
Roof Trussing, Hogged Ridge.... 238
Roof Trussing, King-Post, Effect
of Bad Framing on 237
Roofs, United States 55
Roof, Vertical Pressure of Wind
on, Effect of. 194
Roof, Snow, Weight per Horizon-
tal Foot 193
Roof Weight on Rafter, Effect of.. 179
Roof, Wind, Horizontal and Verti-
cal Pressure of 193
Roofing, Weight of Horizontal and
Inclined 190
Roofing, Weight per Superficial
Foot 190
Roots, Cubes, and Squares, Table
of 638-645
Round Post, Area of 90
Rubble Walls 48
Rulers and Pencil in Drawing.... 540
Rupture Compared with Flexure. 84
Rupture, Crushing, Safe Weight.. 89
Rupture, Sliding, Safe Weight. ... 87
Rupture, Transverse, Safe Weight. 86
Rusting Iron Framing Straps, To
Prevent 239
Safe Load for Material 81
Safe Weight, Allowance for 84
Safe Weight at Any Point, Rule. . 106
Safe Weight, Beam at Middle 103
Safe Weight, Bending 91
Safe Weight, Beam, Breadth of,
To Find 104
Safe Weight, Beam, Depth, To
Find 104
Safe Weight, Breadth or Depth, To
Find, Load at Middle 106
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INDEX.
PAGE
247
240
247
Stairs, Space for Timber and Plas-
ter
Stairs, Stone, Public Building
Stairs, String of, To Make
Stairs, Tread, To Find, Rise Given
242, 246
Stairs, Tread and Riser Connec-
tion 248
Stairs, Width, Rule for 241
Stairs, Winding, Balusters in
Round Rail 313
Stairs, Winding, Bevels in Splayed
Work ... 314
Stairs, Winding, Blocking Out
Rail.. 301
Stairs, Winding, Butt-joint on
Handrail 303
Stairs, Winding, Butt-joint, Cor-
rect Lines for 307
Stairs, Winding, Diagrams Ex-
plained .. 263
Stairs, Winding, Face Mould, Ac-
curacy of 295
Stairs, Winding, Face Mould,
Application 297
Stairs, Winding, Face Mould,
Care in Drawing 295
Stairs, Winding, Face Mould,
Curves Elliptical 301
Stairs, Winding, Face Mould
for 290, 293
Stairs, Winding, Face Mould,
Round Rail 303
Stairs, Winding, Face Mould for
Twist \ . . 291
Stairs, Winding, Flyers and
Winders 251
Stairs, Winding, Front String,
Grade of 253
Stairs, Winding, Handrailing. 256, 289
Stairs, Winding, Handrailing, Bal-
usters Under Scroll 310
Stairs, Winding, Handrailing,
Centres for Square 368
Stairs, Winding, Handrailing,
Face Mould for Scroll 311
Stairs, Winding, Handrailing, Fall-
ing Mould for Raking Scroll. . . 310
Stairs, Winding, Handrailing, Gen-
eral Considerations 258
Stairs, Winding, Handrailing,
Scrolls for 308
Stairs, Winding, Handrailing,
Scroll Over Curtail Step 309
Stairs, Winding,, Handrailing,
Scroll for Curtail Step 310
Stairs, Winding, Scroll at Newel. 309
Stairs, Winding, Illustrations by
Planes 261
Stairs, Winding, Moulds for
Quarter Circle 255
Stairs, Winding, Newel Cap, Form
of 312
Stairs, Winding, Objectionable. . 240
Stairs, Winding, Pitch Board, To
Obtain 252
Stairs, Winding, Rise and Tread,
To Obtain 251
Stairs, Winding, Sliding of Face
Mould. 299
Stairs, Winding, String, To Ob-
tain 252
Stairs, Winding, Timbers, Posi-
tion of 252
Stairs and Windows, How. Ar-
ranged 42
Stiles of Windows, Allowance for. 319
St. Mark, Tenth or Eleventh Cen-
tury 12
Stone Bridge Building, Truss
Work 232
Stone Bridge, Building Arch.... 230
Stone Bridge, Centres for, Con-
struction 229
Stone Bridge, Pressure on Arch
Stones 230
Stop for Doors 317
Stores, Floor Beams, Safe Weight. 126
Stores, Ordinary, Floor-Beams,
Sizes, To Find 127
Stores, First-Class, Floor-Beams,
Sizes, To Find 128
St. Paul's, London, Dome of 54
St. Peter's, Rome, Fourteenth and
Fifteenth Centuries 12
Straight or Right Line Defined. . . 544
INDEX.
681
Strains, Cross, Resistance to 77
Strains on Domes, Tendency of. . 219
Strains Exceed Weights 61
Strains, Graphic Representation.. 165
Strain Greatest at Middle of Beam. 105
Strains by Parallelogram of
Forces 165
Strains, Practical Method cf De-
termining 62
Strains of Rafter in Roof 205
Strains, Resistance, To What Due. 78 |
Strain on Roof Timbers Shown
Geometrically 190
Strains on Roof Timbers Geomet-
rically Applied 202 j
Strains on Roof Timbers, Parallel-
ogram of Forces 198 ;
Strain, Shearing, Tubular Iron
Girder 157
Strain Unequal, Cause of 83
Straps, Iron, Roof Truss 239 j
Strassburg, Cathedral of 12 j
Strassburg, Towers of the Min-
ster II I
Strength and Stiffness of Mate-
rials..- ."... 78 j
Structure of Materials 76 |
Struts Denned 173 j
Struts and Ties 68 i
Struts and Ties, Difference Be-
tween 69 :
St. Sophia, Sixth Century 12 i
Stucco Cornice for Interior 340 ;
Studs and Joists Defined 174 '
Styles, Grecian, Only Known by
Them 16 !
Stylobate, Substructure 'for Col-
umns 14 •
Subnormal and Normal (in Para-
bola) 496 I
Subtangent, Parabola 496
Subtangent of Ellipse Defined... 486
Subtraction and Addition (Alge-
bra) 398
Superficies Defined (Geometry)... 544
Supports, Girders, Length, Rule.. 157
Supports, Position of 65
Supports, Inclination of, Unequal. 60
Suspension Rods, Location in
Roof 212
Suspension Rods in Roof, Safe
Weight 210
Symbols Chosen at Pleasure (Al-
gebra) 395
Symbols, Transferring (Algebra). 399
Systyle, Explanation of 20
Table of Circles 649-652
Table of Contents 6^3-624
Table of Capacity of Wells, Cis-
terns, etc 653
Table of Squares, Cubes, and
Roots 638-645
Table of Woods, Description of. . 80
Tail-Beams Defined 130
Tanged Curve, To Describe (Ge-
ometry) 565
Tangent to Axes, Ellipse . . . . 485
Tangent Defined 547
Tangent with Foci, Ellipse 487
Tangent to Ellipse. To Draw 592
Tangent at Given Point in Cir-
cle 557
Tangent at Given Point, Without
Centre 557
Tangent of Parabola 493
Tangents and Sines, Logarithms
(Polygons) 464
Temples Built in the Doric Style. 19
Temple, Doric, Origin of the 17
Temple of Jupiter at Thebes 33
Tenons and Splices, Knowledge
Important 88
Tensile Strain, Area of Piece, To
Find 99
Tensile Strain, Compressed Ma-
terial ioo
Tensile Strain, Condition of Sus-
pended Piece 98
Tensile Strain, Safe Weight 96
Tensile Strain, Safe Weight, To
Compute 97
Tensile Strain, Sectional Area, To
Obtain 97
Tensile Strain, Suspended, M»te-
terial Extended ioo
682
INDEX.
PAGE
Tensile Strain on Tie-Beam in Roof
Truss 204
Tensile Strain, Weight of Suspend-
ed Piece 98
Tensile Strength of Cast Iron 161
Tension and Compression, Framed
Girders 174
Tension, Resistance to 77
Tension, Resistance to, Table of
Materials 82
Tension, Resistance to, Results
Obtained 82
Tension, Resistance to, Proportion
in Materials 81
Tent, Habitation of the Shepherd. 13
Testing Machine, Description in
Transverse Strains 80
Testing Materials, Hydraulic
Method 80
Testing Materials, Dates of 80
Testing Materials, Manner of. ... 80
Tetragon Defined 546
Tetragon, Radius of Circles
(Polygons) 446
Tetrastyle, Intercolumniation 20
Thebes, Thickness of Walls at. . . 33
Thrust, Horizontal 63
Thrust, Horizontal, Examples. ... 64
Thrust, Horizontal, Tendency of.. 88
Tie-Beam in Ceiling, Load on... 190
Tie-Beam and Rafter, Safe Weight. 87
Tie-Beam in Roof 54
Tie-Beam in Roof, Tensile Strain. 204
Tie-Rods, Diameter, To Find 164
Tie-Rods, Floor Arches, Dwell-
ings 153
Tie-Reds, Floor Arches, First-
Class Stores 153
Tie-Rods, Wrought Iron 164
Ties Defined 173
Ties and Struts, To Distinguish.. 69
Ties and Struts, Framed Girders.. 174
Ties and Struts, Principles of. ... 68
Ties, Timbers in a State of Ten-
sion 68
Titus, Composite Arch of 28
Trimme/, Breadth, To Find, Two
Sets Tail-Beams 134
Top Rail, Doors, Width, Rule 317
Torus, Classic Moulding 323
Torus, Grecian Moulding. 326
Tower of Babel, History of 5
Towers of the Minster, Strassburg. n
Transverse Axis Defined 548
Transverse Strains, Compressed
and Extended, Material 100
Transverse Strains, Defined 99
Transverse Strains, Explanation
of Table III 101
Transverse Strains, Greater
Strength of One Piece 101
Transverse Strains, Neutral Line
Defined 100
Transverse Strains, Proportion to
Breadth 101
Transverse Strains, Hatfield's,
Reference to. .80, 121, 133, 138,
143, 144, 145, 146, 148
Transverse Strains, Resistance to,
Table of 83
Transverse Strains, Description of
Table 84
Transverse Strains, Strength Di-
minished by Division 101
Trapezoid Defined 546
Trapezium Defined 546
Tread, To Find, Rise Given
(Stairs) 242,246
Tread and Nosing, Position of
(Stairs) 241
Tread and Rise, To Find, Winding
Stairs 251
Tread and Rise, To Find, Blon-
del's Method 242
Tread and Rfcse, Table for Shops \
and Dwellings 245
Tread and Riser, Connection of
(Stairs) 248
Triangle, Altitude of (Polygons). 442
Triangles, Base, Formula for (Trig-
onometry) 516
Triangle, Construct a (Geometry). 587
Triangle, Construct Equal-Sided
(Geometry) 575
Triangle Defined 545
Triangle, Examples (Geometry).. . 350
INDEX.
683
PAGE
Triangles, Equal Altitude 361
Triangle Equal Quadrangle 353
Triangles, Equation of (Trigo-
nometry) 515
Triangles, Homologous (Geom-
etry) 362
Triangles, Hypothenuse, Formula
for 516
Triangles, Hypothenuse, To Find
(Trigonometry) 518
Triangles, Perpendicular, To Find
(Trigonometry) 517
Triangle or Set-Square in Draw-
ing 539
Triangle or Set-Square, Use of. . . 541
Triangles, Terms Denned (Trigo-
nometry) 512
Triangles, Three Angles Equal
Right Angle 354
Triangles, Value of Sides (Trigo-
nometry) 516
Trigon, Irregular Polygons (Ge-
ometry) 546
Trigon, Radius of Circle (Poly-
gons) 443
Trigon, Rule (Polygons) 441
Trigonometry, Oblique Triangles,
Two Angles 523
Trigonometry, Oblique Triangles,
Two Sides 521
Trigonometry, Oblique Triangles,
First Class 520
Trigonometry, Oblique Triangles,
Second Class 522
Trigonometry, Oblique Triangles,
Third Class 526
Trigonometry, Oblique Triangles,
Fourth Class
Trigonometry, Oblique Triangles,
Sines and Sides
Trigonometry, Oblique Triangles,
Formula, First Class 531
Trigonometry, Oblique Triangles,
Formula, Second Class 532
Trigonometry, Oblique Triangles,
Formula, Third Class 534
Trigonometry, Oblique Triangles,
Formula, Fourth Class 534
S*\
519
Trigonometry, Right Angled Tri-
angles 510
Trigonometry, Right Angled Tri-
angles, Third Side, To Find... 511
Trigonometry, Right Angled Tri-
angle, Formula 530
Trigonometry, Tables 513
Trigonometry, Triangles, Base,
Formula for 516
Trigonometry, Triangles, Equa-
tions of 515
Trigonometry, Triangles, Hypoth-
enuse, Formula 516
Trigonometry, Triangles, Hypoth-
enuse, To Find 518
Trigonometry, Triangles, Perpen-
dicular, To Find 517
Trigonometry, Triangles, Terms
Denned 512
Trigonometry, Triangles, Value of
Sides 516
Trimmer or Carriage Beam,
Breadth, To Find 132
Trimmer or Carriage Beams De-
nned 130
Trimmer, One Header, Breadth,
To Find, Dwellings and Stores. 133
Trimmer, Well-Hole in Middle,
Breadth, To Find 136
Trisect a Right Angle 5^4
Truss, Diagram of. 200
Truss, Force Diagrams, Figs. 59,
68 and 69 179
Figs. 60, 70 and 71 iSo
Figs. 61, 72 and 73 181
Figs. 63, 74 and 75 183
Figs. 64, 77 and 78 184
Figs. 65, 78 and 79 185
Figs. 66, 80 and 81 186
Truss, Roof, Framing for 237
Truss, Roof, Iron Straps 239
Truss, Weight, per Horizontal
Foot, To Find 192
Truss Work, Stone Bridge Build-
ing 232
Trussing and Framing, Gravity
and Resistance 76
Trussing Partitions, Effect of 175
684
INDEX.
Trussing Roofs, Effect of 178
T-Square, How to Make 539
Tubular Iron Girder, Area of Bot-
tom Flange, Dwellings 159
Tubular Iron Girder, Area of Bot-
tom Flange, First-Class Stores. 160
Tubular Iron Girder, Arc of
Flange, Load at Middle 154
Tubular Iron Girder, Area of
Flange, Load Any Point 155
Tubular Iron Girder, Area of
Flange, Load Uniform 156
Tubular Iron Girder, Flanges,
Proportion of. 157
Tubular Iron Girder, Construction
of 154
Tubular Iron Girder, Rivets, Al-
lowance for 157
Tubular Iron Girder, Shearing
Strain 157
Tubular Iron Girders, Web of. . . 158
Tuscan, Modern, Appropriate for
Buildings 30
Tuscan Order, Introduction of the. 30
Tuscan Order, Principal Style in
Italy 30
Twelfth Century, Buildings in the. n
Twist Rail, Platform Stairs 277
Twists, Stairs, Nicholson's Method
for 259
Undecagon Defined
United States, Roofs in.
Vault, Simple, Groined or Com-
plex
Ventilation, Proper Arrangement
for...!.
Versed Sine of Arc, To Find
Vertical Pressure of Wind on
Roof.
Vertical Tangent of Parabola De-
fined
Volutes, To Describe the
Voussoir of an Arch. . ,
Wall, The...
Walls, Coffer.
546
55
52
45
56i
194
495 !
20 *
52 I
|
48 I
49!
PAGE
Walls, Construction and Forma-
tion 48
Walls, Eddystone and Bell Rock
Lighthouses 48
Walls, Egyptian, Massiveness of.. 33
Walls, Modern Brick 49
Walls of Pantheon and Roman
Buildings 49
Walls of Pantheon at Rome 53
Walls, Pise, of France 49
Walls, Reticulated 49
Walls, Rubble -. ... 48
Walls, Strength of. 48
Walls, Various Kinds 49
Walls, Wooden 49
Weakening Girder, Manner of. . . 140
Web of Tubular Iron Girder,
Thickness of 158
Weight of Materials for Building
Table of 654-656
Wells, Cisteins, etc., Table of
Capacity 653
White Pine, Weights of Beams
Table of 177
Wind, Greatest Pressure, per Su-
perficial Foot 90
Wind on Roof, Effect of Vertical
Pressure 194
Wind on Roof, Horizontal and
Vertical Pressure 193
Winders in Stairs, How to Place
the 42
Winders and Flyers, Stairs 251
Windows, Arrangement of 44
Windows, Circular Headed 320
Windows, Circular Headed, To
Form Soffit 321
Windows, Dimensions, To Find. 318
Window-Frame, Size of 318
Windows, Front of Building, Ef-
fect of 320
Windows, Heights, Table of,
Width Given 320
Windows, Height from Floor. . . . 320
Windows, Inside Shutters, Re-
quirement 319
Windows, Position and Light
from.. 317
INDEX.
685
Windows and Stairs, How Ar-
ranged 42
Windows, Stiles, Allowance lor.. 319
Windows, Width Uniform, Height
Varying . . . : 319
Winding Stairs, Balusters in
Round Rail 313
Winding Stairs, Bevels in Splayed
Work 314
Winding Stairs, Blocking Out
Rail.. 301
Winding Stairs, Butt Joint, Posi-
tion of. 303
Winding Stairs, Butt Joint 307
Winding Stairs, |Diagram of, Ex-
plained 263
Winding Stairs, Face Mould for
290,293
Winding Stairs, Face Mould, Ac-
curacy of. 295
Winding Stairs, Face Mould, Ap-
plication of 297
Winding Stairs, Face Mould,
Curves Elliptical 301
Winding Stairs, Face Mould,
Drawing 296
Winding Stairs, Face Mould,
Round Rail 303
Winding Stairs, Face Mould, Slid-
ing of. 299
Winding Stairs, Face Mould for
Twist 291
Winding Stairs, Flyers and Wind-
ers 251
Winding Stairs, Front String,
Grade of. 253
Winding Stairs, Handrailing
. ; 256, 289
Winding Stairs, Handrailing, nal-
usters Under Scroll 310
Winding Stairs, Handrailing, Cen-
tres in Square 308
Winding Stairs, Handrailing, Face
Mould for Scroll. .., 311
Winding Stairs, Handrailing, Fall-
ing Mould. . . 310
Winding Stairs, Handrailing,
General Considerations 258
Winding Stairs, Handrailing,
Scrolls for 308
Winding Stairs, Handrailing,
Scroll Over Curtail Step 309
Winding Stairs, Handrailing,
Scroll for Curtail Step 310
Winding Stairs, Handrailing,
Scrolls at Newel 309
Winding Stairs, Illustrations by
Planes 261
Winding Stairs, Moulds for Quar-
ter Circle 255
Winding Stairs, Newel Cap, Form
of 312
Winding Stairs Objectionable. . . . 240
Winding Stairs, Pitch Board, To
Obtain 252
Winding Stairs, Rise and Tread,
To Obtain 251
Winding Stairs, String, To Obtain. 252
Winding Stairs, Timbers, Posi-
tion of. 252
Wood, Destruction by Fire 37
Wooden Beams, Use Limited 154
Woods, Hydraulic Method of
Testing 80
Wreath for Round Rail, Platform
Stairs 267
THE END.
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